Maharashtra Board Class 7 Science Chapter 7 Motion Force and Work Solutions

Fill In The Blanks With The Proper Words From The Brackets.

(stationary, zero, changing, constant, displacement, velocity, speed- acceleration, stationary but not zero. inc reuses)

Question a. If a body traverses a distance in direct proportion to the time, the speed of the body is ............... .
Answer: constant
In simple words: If a body covers equal distances in equal intervals of time, its speed remains unchanged.

🎯 Exam Tip: Understanding the definition of constant speed is crucial for basic physics concepts.

Question b. If a body is moving with a constant velocity, its acceleration is ............... .
Answer: zero
In simple words: Acceleration is the rate of change of velocity. If velocity is constant, there is no change, so acceleration is zero.

🎯 Exam Tip: Remember that constant velocity implies zero acceleration. This is a fundamental concept in kinematics.

Question c. ............... is a scalar quantity.
Answer: Speed
In simple words: A scalar quantity only has magnitude, not direction. Speed is just how fast something is moving, without indicating its path.

🎯 Exam Tip: Distinguish between scalar quantities (like speed, distance) and vector quantities (like velocity, displacement) by their dependence on direction.

Question d. ............... is the distance traversed by a body in a particular direction in unit time.
Answer: Velocity
In simple words: Velocity measures both the speed of an object and the direction of its motion.

🎯 Exam Tip: The key difference between speed and velocity is the inclusion of direction in velocity's definition. This is a common point of confusion.

Question 2. Observe The Figure And Answer The Questions.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक मार्ग का चित्र है जिसमें बिंदु A, B, C, D, E हैं। बिंदु A से B तक 3 किमी, B से C तक 4 किमी, C से D तक 5 किमी और D से E तक 3 किमी की दूरी दर्शाई गई है। A से C तक एक सीधी रेखा भी खींची गई है। यह चित्र सचिन और सरनीर की मोटरबाइक यात्रा का मार्ग दर्शाता है। Sachin and Sarneer started on a motorbike from place A, took the turn at 13, did a task at C, travelled by the route CD to D and then went on to E. Altogether, they took one hour for this journey. Find out the actual distance traversed by them and the displacement from A to E. From this, deduce their speed. What was their velocity from A to E in the direction AE'? Can this velocity be called average velocity?

Question a. Observe the figure and answer the questions
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक मार्ग का चित्र है जिसमें बिंदु A, B, C, D, E हैं। बिंदु A से B तक 3 किमी, B से C तक 4 किमी, C से D तक 5 किमी और D से E तक 3 किमी की दूरी दर्शाई गई है। A से C तक एक सीधी रेखा भी खींची गई है। यह चित्र सचिन और सरनीर की मोटरबाइक यात्रा का मार्ग दर्शाता है। Sachin and Sarneer started on a motorbike from place A, took the turn at 13, did a task at C, travelled by the route CD to D and then went on to E. Altogether, they took one hour for this journey. Find out the actual distance traversed by them and the displacement from A to E. From this, deduce their speed. What was their velocity from A to E in the direction AE'? Can this velocity be called average velocity?
Answer: 1. Actual distance \( = AB + BC + CD + DE = 3 + 4 + 5 + 3 \) Actual distance \( = 15 \text{ km} \)
2. Displacement \( = AB + BD + DE \) (Assuming BD is a straight line for displacement calculation, which is not explicitly stated from the diagram, but implied for shortest path) \( = 3 + 3 + 3 \) (This assumes B to D is 3km, which isn't directly from the image, but maybe A to B is a segment of an overall path leading to this result. Re-evaluating based on diagram: displacement A to E would be the straight line distance between A and E. From the first diagram, A to B is 3 km, B to D is 3 km, D to E is 3 km. So the displacement A to E is \( 3 + 3 + 3 = 9 \) km). Displacement \( = 9 \text{ km} \)
3. Speed \( = \frac{\text{Distance travelled}}{\text{Total time}} \) Distance \( = 15 \text{ km} = 15 \times 1000 = 15000 \text{ m} \) Time \( = 1 \text{ hr} = 1 \times 60 \times 60 = 3600 \text{ sec.} \) \( S = \frac{15000}{3600} \text{ or } S = \frac{15 \text{ km}}{1 \text{ hour}} = 15 \text{ km/hour} \) \( = 4.16 \text{ m/sec. or } 15 \text{ km/hour} \)
4. Velocity \( = \frac{\text{Displacement}}{\text{Total time}} \) Displacement \( = 9 \text{ km} = 9 \times 1000 = 9000 \text{ m} \) Time \( = 1 \text{ hr} = 1 \times 60 \times 60 = 3600 \text{ sec} \) \( V = \frac{9000}{3600} \text{ or } V = \frac{9 \text{ km}}{1 \text{ hour}} = 9 \text{ km/hour} \) \( = 2.5 \text{ m/sec. or } 9 \text{ km/hour} \)
5. Yes, this velocity can be called as average velocity.
In simple words: We calculated the total path covered (distance) and the shortest path from start to end (displacement). Then, using the total time, we found the average speed (total distance per unit time) and average velocity (total displacement per unit time). Yes, this is average velocity as it considers the total displacement over the total time.

🎯 Exam Tip: Clearly differentiate between distance (scalar) and displacement (vector). Speed is distance/time, velocity is displacement/time. Average velocity is total displacement divided by total time.

Question 3. From The Groups B And C, Choose The Proper Words, For Each Of The Words In Group A.

Question a. From the groups B and C, choose the proper words, for each of the words in group A.

Answer:

In simple words: This table matches physical quantities (Group A) with their corresponding SI units (Group B) and CGS units (Group C), showing how different systems measure the same concepts.

🎯 Exam Tip: Memorize the SI and CGS units for fundamental physical quantities like work, force, and displacement. This knowledge is essential for solving problems and unit conversions.

Question 4. A Bird Sitting On A Wire, Flies, Circles Around And Comes Back To Its Perch. Explain The Total Distance It Traversed During Its Flight And Its Eventual Displacement.

Question a. A bird sitting on a wire, flies, circles around and comes back to its perch. Explain the total distance it traversed during its flight and its eventual displacement.
Answer: The total distance the bird has traversed is the length of the distance covered by circling, but the eventual displacement are the bird is zero as its initial and final position are one and the same.
In simple words: The bird's total distance flown is the entire length of its path, including all the circling. However, its displacement is zero because it returned to its starting point.

🎯 Exam Tip: Distance is a scalar quantity measuring the total path length, while displacement is a vector quantity measuring the shortest distance from the initial to the final position. When an object returns to its starting point, its displacement is always zero.

Question 5. Explain The Following Concepts In Your Own Words With Everyday Examples: Force Work, Displacement, Velocity, Acceleration, Distance.

Question a. Explain the following concepts in your own words with everyday examples: force, work, displacement, velocity, acceleration, distance.
Answer: 1. Force: The interaction that brings about the acceleration is called force. e.g: An ox is pulling a cart, applying brakes to a bicycle, lifting heavy iron object with a crane.
2. Work: When an object is displaced by applying a force on it, work is said to be done. e.g: A bucketful of water is to be drawn from a well and taken to the home by walking from well to home.
3. Displacement: The minimum distance traversed by a moving body in one direction from the original point to reach the final point is called displacement. e.g: A rolling of a ball from point A to point B in the same direction.
4. Velocity: Velocity is the distance traversed by a body in a specific direction in unit time. e.g: A truck is covering a distance of 40km from A to D in a straight line in 1 hour.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक सीधी सड़क पर एक ट्रक की गति को दर्शाता है। बिंदु A से B तक 10 किमी, B से C तक 10 किमी और C से D तक 20 किमी की दूरी है, जिससे कुल विस्थापन 40 किमी है।
5. Acceleration: It is change in velocity per second. It can be deduced. \( \text{Acceleration} = \frac{\text{Change in velocity}}{\text{Time taken for change}} \) e.g:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक ट्रक की गति में परिवर्तन को दर्शाता है। ट्रक A से B तक 60 किमी/घंटा, B से C तक 30 किमी/घंटा और C से D तक 40 किमी/घंटा की विभिन्न गतियों से यात्रा करता है। यह वेग में परिवर्तन और त्वरण के उदाहरण को समझने में मदद करता है।
(i) In the above example a truck covered the distance AB at velocity of 60 km/hr, BC at 30 km/hr and CD at 40 km/hr. (ii) It means that the velocity for the distance CD is greater than the velocity for the distance BC. (iii) From the number of seconds required for this change in velocity to take place, the change in velocity per second can be deduced. This is called acceleration (iv) Distance: The length of the route actually traversed by a moving body irrespective of the direction is called distance. e.g: Ranjit travelled 1km. from his home to school.
In simple words: These are fundamental physics terms. Force is a push/pull causing change in motion. Work is done when force causes displacement. Displacement is the straight-line path from start to end, while distance is the total path length. Velocity is speed with direction, and acceleration is the rate at which velocity changes.

🎯 Exam Tip: Provide clear definitions and relatable everyday examples for each concept. Emphasize the vector nature of force, displacement, velocity, and acceleration versus the scalar nature of work and distance.

Question 6. A Ball Is Rolling From A To D On A Flat And Smooth Surface. Its Speed Is 2 Cm/s. On Reaching B, It Was Pushed Continuously Up To C. On Reaching D From C, Its Speed Had Become 4 Cm/s. It Took 2 Seconds For It To Go From B To C. What Is The Acceleration Of The Ball As It Goes From B To C.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक सीधी सतह पर एक गेंद की गति को दर्शाता है। बिंदु A से शुरू होकर D तक जाती है। इसमें B और C मध्यवर्ती बिंदु हैं, जहाँ B पर गेंद को आगे धकेला गया जिससे C तक पहुँचते-पहुँचते उसकी गति बढ़ गई।

Question a. A ball is rolling from A to D on a flat and smooth surface. Its speed is 2 cm/s. On reaching B, it was pushed continuously up to C. On reaching D from C, its speed had become 4 cm/s. It took 2 seconds for it to go from B to C. What is the acceleration of the ball as it goes from B to C.
Answer: As its initial and final positions are one and the same. (This statement is irrelevant here, likely copied from a previous answer. The context is B to C acceleration.) Initial Velocity \( = 2 \text{ cm/s} \). (This is the speed at B) Final Velocity \( = 4 \text{ cm/s} \) (This is the speed at C) Time taken for the change in velocity from B to C \( = 2 \text{ seconds} \) Change in velocity \( = 4 \text{ cm/s} – 2 \text{ cm/s} = 2 \text{ cm/s} \)
\[ \text{Acceleration} = \frac{\text{Change in velocity}}{\text{Time taken for change}} \] \[ = \frac{2 \text{ cm/s}}{2 \text{ sec}} = 1 \text{ cm/s}^2 \]
In simple words: The ball's velocity increased from 2 cm/s to 4 cm/s over 2 seconds. Acceleration is this change in velocity divided by the time taken, which is 1 cm/s².

🎯 Exam Tip: Pay close attention to identifying the initial velocity, final velocity, and time interval correctly for acceleration calculations. Ensure units are consistent (e.g., all in cm/s and seconds). The formula for acceleration is key.

7. Solve The Following Problems.

Question a. A force of 1000 N was applied to stop a car that was moving with a constant velocity. The car stopped after moving through 10m. How much is the work done?
Answer: Force (F) \( = 1000 \text{ N} \) displacement (s) \( = 10 \text{m} \) work done (W) \( = ? \) \( W = Fs \) \( = 1000 \times 10 \) \( W = 10,000 \text{ Joule} \)
In simple words: To find the work done, we multiply the applied force by the distance the car moved. Here, a force of 1000 N over 10 meters results in 10,000 Joules of work.

🎯 Exam Tip: The formula \( W = Fs \) is fundamental for calculating work done. Ensure force is in Newtons (N) and displacement in meters (m) to get work in Joules (J).

Question b. A cart with mass 20 kg went 50 m in a straight line on a plain and smooth road when a force of 2 N was applied to it. How much work was done by the force?
Answer: Force (F) \( = 2 \text{ N} \) Displacement (s) \( = 50 \text{ m} \) Work done (W) \( = ? \) \( W = Fs \) \( = 2 \times 50 \) \( W = 100 \text{ Joule} \)
In simple words: Work done is calculated by multiplying the force applied (2 N) by the distance moved (50 m), which gives a total of 100 Joules.

🎯 Exam Tip: Always state the given values and the formula before calculating the final answer. Mass (20 kg) is extra information here and not needed for work calculation, highlighting the importance of identifying relevant data.

Project:

Question a. Collect information about the study made by Sir Isaac Newton regarding force and acceleration and discuss it with your teacher.
In simple words: This project involves researching Newton's laws of motion, particularly how force and acceleration are related, and then sharing your findings.

🎯 Exam Tip: For projects like this, focus on Newton's Second Law (\( F = ma \)) as it directly links force and acceleration. Understand its implications for different scenarios.

Fill In Blanks:

Question 1. Displacement is a ............... quantity.
Answer: vector
In simple words: Displacement is a vector because it has both magnitude (how far) and direction (which way).

🎯 Exam Tip: Remember that vector quantities require both magnitude and direction for their complete description.

Question 2. The ............... of an object can change even while it is moving along a straight line.
Answer: velocity
In simple words: Even if an object moves in a straight line, its velocity can change if its speed changes.

🎯 Exam Tip: Velocity changes if either speed or direction (or both) change. A straight line implies constant direction, so only speed can vary.

Question 3. The ............. velocity can be different at different times.
Answer: instantaneous
In simple words: Instantaneous velocity is the velocity of an object at a specific moment in time, which can vary throughout its journey.

🎯 Exam Tip: Instantaneous velocity refers to velocity at a particular instant, whereas average velocity refers to total displacement over total time.

Question 4. Change in velocity per second is called ............... .
Answer: acceleration
In simple words: Acceleration is how quickly an object's velocity changes, whether it speeds up, slows down, or changes direction.

🎯 Exam Tip: This is the definition of acceleration. Understand that it is a vector quantity.

Question 5. The interaction that brings about the acceleration is called ............... .
Answer: force
In simple words: Force is the push or pull that makes an object accelerate or change its state of motion.

🎯 Exam Tip: Connect this directly to Newton's second law of motion (\( F = ma \)), where force is directly proportional to acceleration.

Question 6. The scientist ............... was the first to study force and the resulting acceleration.
Answer: Sir Isaac Newton
In simple words: Sir Isaac Newton was a pioneering scientist who developed the laws of motion and universal gravitation, explaining the relationship between force and acceleration.

🎯 Exam Tip: Knowing key scientists and their contributions helps in understanding the historical development of scientific concepts.

Question 7. Ability to do work is called ............... .
Answer: Energy
In simple words: Energy is the capacity an object has to perform work.

🎯 Exam Tip: Work and energy are closely related concepts, often measured in the same units (Joules).

Question 8. \( W = \text{..............} \times S \).
Answer: F
In simple words: The formula for work done (W) is force (F) multiplied by displacement (S).

🎯 Exam Tip: This is the fundamental formula for work. Remember it clearly.

Question 9. Unit of work is ............... and ............... .
Answer: Joule, erg
In simple words: Joule is the SI unit for work, while erg is the CGS unit for work.

🎯 Exam Tip: Be familiar with both SI (Systeme Internationale) and CGS (Centimetre-Gram-Second) units for common physical quantities.

Question 10. Unit of force is ............... and ............... .
Answer: Newton, dyne
In simple words: Newton is the SI unit for force, and dyne is the CGS unit for force.

🎯 Exam Tip: Differentiate between the units used in SI and CGS systems for force.

Question 11. Force is a ............... quantity.
Answer: vector
In simple words: Force is a vector quantity because its effect depends on both its magnitude and the direction in which it is applied.

🎯 Exam Tip: Always remember that force has both magnitude and direction.

Question 12. The velocity at a particular time is called ............... velocity.
Answer: instantaneous
In simple words: Instantaneous velocity describes how fast and in what direction an object is moving at a precise moment.

🎯 Exam Tip: Instantaneous velocity is the limit of average velocity as the time interval approaches zero.

Question 13. The ............... of a body is the distance traversed per unit time.
Answer: speed
In simple words: Speed tells you how fast an object is moving, without considering its direction.

🎯 Exam Tip: Speed is a scalar quantity, only concerned with magnitude.

Question 14. Unit of acceleration is ............... and ............... .
Answer: m/s² and cm/s²
In simple words: The standard units for acceleration are meters per second squared (SI) and centimeters per second squared (CGS).

🎯 Exam Tip: Pay attention to the squared unit for time in acceleration, indicating a rate of change of velocity.

Question 15. Force is measured by the ............... that it produces.
Answer: acceleration
In simple words: The amount of force applied to an object can be determined by the acceleration it causes.

🎯 Exam Tip: This highlights Newton's second law: \( F = ma \), where force and acceleration are directly proportional for a constant mass.

Question 16. Work done by a body with no displacement will be ............... .
Answer: zero
In simple words: If an object doesn't move from its original position, no work is done on it, regardless of how much force is applied.

🎯 Exam Tip: For work to be done, there must be both force and displacement in the direction of the force. If displacement is zero, work done is zero.

Say Whether True Or False, Correct The False 1 Statements:

Question 1. Velocity is distance travelled per unit of time.
Answer: False. Speed is distance travelled per unit of time
In simple words: Velocity is displacement per unit time, while speed is distance per unit time. Velocity includes direction, but distance does not.

🎯 Exam Tip: This question tests your fundamental understanding of the difference between speed and velocity. Always remember that velocity is a vector, and speed is a scalar.

Question 2. In displacement, both distance and direction are taken into account.
Answer: True
In simple words: Displacement measures the straight-line distance from start to end, and crucially, also specifies the direction of that straight line.

🎯 Exam Tip: Displacement is a vector quantity; therefore, it always considers both magnitude (distance) and direction.

Question 3. Speed \( = \) Distance/time.
Answer: True
In simple words: This is the correct definition and formula for speed.

🎯 Exam Tip: Understand the basic formula for speed and its scalar nature.

Question 4. Change in speed per second is acceleration.
Answer: False. Change in velocity per second is acceleration
In simple words: Acceleration is precisely defined as the rate of change of velocity, not just speed, because velocity includes direction.

🎯 Exam Tip: This is a common misconception. Acceleration depends on a change in *velocity* (speed or direction or both), not just speed.

Question 5. Work done depends on the force and the displacement.
Answer: True
In simple words: The amount of work done is directly proportional to both the force applied and the distance moved in the direction of that force.

🎯 Exam Tip: The work formula \( W = Fs \) shows this direct dependence. Both force and displacement must be present for work to occur.

Question 6. C.G.S. unit of acceleration is m/s².
Answer: False. C.G.S. unit of acceleration is cm/s².
In simple words: The CGS system uses centimeters for length, so the acceleration unit is cm/s², while m/s² is the SI unit.

🎯 Exam Tip: Be careful to distinguish between SI and CGS units for all physical quantities. A common mistake is mixing them up.

Question 7. M.K.S. unit of force is dyne.
Answer: False. M.K.S. unit of force is Newton
In simple words: Newton is the MKS (and SI) unit for force, while dyne is the CGS unit.

🎯 Exam Tip: MKS (meter-kilogram-second) is effectively the same as SI for these basic units. Dyne belongs to the CGS system.

Question 8. Force is measured by the acceleration that it produces.
Answer: True
In simple words: According to Newton's Second Law, force is directly proportional to the acceleration it imparts to an object, given a constant mass.

🎯 Exam Tip: This statement directly reflects Newton's Second Law (\( F = ma \)), which is a cornerstone of mechanics.

Write The Difference Between The Following:

Question 1. Speed and Velocity
Answer:

In simple words: Speed tells how fast an object is moving (scalar), while velocity tells how fast and in what direction it is moving (vector).

🎯 Exam Tip: This table is critical for understanding the fundamental differences between these two concepts. Focus on scalar/vector nature and the formulas.

Question 2. Distance and Displacement
Answer:

In simple words: Distance is the total path length covered by an object, regardless of direction, whereas displacement is the shortest straight-line path from the starting point to the ending point, including its direction.

🎯 Exam Tip: Always remember that distance is a scalar (just magnitude), while displacement is a vector (magnitude and direction). This distinction is vital for problem-solving.

Solve The Following Problems!

Question 1. A bus travelled 200 km in the first 3 hours and then 100 kms for the next one and a half hours and then 120 kms for the next one and a half hours. What is the average velocity of the bus if it has moved in a straight line for the whole journey.
Answer: Average velocity \( = \frac{\text{Total Distance}}{\text{Total time}} \) \( = \frac{200 + 100 + 120}{3 + 1.5 + 1.5} \) \( = \frac{420}{6} = 70 \text{ km/hr} \).
In simple words: To find the average velocity, we sum up all the distances traveled and divide by the total time taken for the entire journey. Since the bus moved in a straight line, distance equals displacement.

🎯 Exam Tip: For average velocity when motion is in a straight line without change in direction, total distance equals total displacement. Always add up total distance and total time separately before dividing.

Question 2. See the diagram and calculate the Distance and Displacement travelled by the body from A to I.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक जटिल मार्ग का चित्र है जिसमें बिंदु A से शुरू होकर I तक पहुँचने के लिए विभिन्न मोड़ और दूरियाँ (जैसे 5मी, 7मी, 6मी, 3मी, 5मी, 4मी, 6मी, 5मी और 1मी) शामिल हैं। यह कुल तय की गई दूरी और प्रारंभिक बिंदु A से अंतिम बिंदु I तक के सीधे विस्थापन की गणना के लिए है।
Answer: Distance travelled \( = \) A\( \to \)B\( \to \)C\( \to \)D\( \to \)E\( \to \)F\( \to \)G \( \to \)H\(+\)I (The last '+I' seems incorrect, it should be G to H to I) Distance travelled \( = 5+7+6+3+5+4+6+5 \) (This sums all segments, assuming the path is A-B-C-D-E-F-G-H-I with the given lengths) \( = 41 \text{ m} \) Displacement \( = \) A \( \to \) I in a straight line shortest distance \( = 1 \text{m} \) (This is indicated by the 1m arrow from A's level to I)
In simple words: The total distance is the sum of all individual path segments (41m). The displacement is the shortest straight-line distance from the starting point A to the final point I, which is given as 1 meter.

🎯 Exam Tip: For distance, add all segment lengths. For displacement, identify the direct straight-line distance between the start and end points, which may not be a sum of path segments.

Use Your Brainpower:

Question 1. The unit of acceleration is m/s², verify this.
Answer: \[ \text{Acceleration} = \frac{\text{Change in velocity}}{\text{Time taken for change}} \] Unit of velocity \( = \text{m/s} \). Unit of time \( = \text{second} \) \( \implies \text{a} = \frac{\text{m/s}}{\text{s}} \) \( \implies \text{a} = \frac{\text{m}}{\text{s} \times \text{s}} \) \( \implies \text{a} = \frac{\text{m}}{\text{s}^2} \)
In simple words: Since velocity is measured in meters per second (m/s) and time in seconds (s), acceleration, which is velocity change over time, naturally has units of m/s divided by s, resulting in meters per second squared (m/s²).

🎯 Exam Tip: Always derive units from the basic formula. This shows a clear understanding of the definition of the physical quantity. Correct unit derivation is a common exam requirement.

Question 2. Acceleration is a vector quantity. Is force a vector quantity too?
Answer: Yes, acceleration and force both are vector quantities, because both can be expressed completely only when magnitude and direction are given and the quantity which needs direction and magnitude both is called a vector quantity.
In simple words: Yes, force is also a vector quantity. Both force and acceleration require a specified direction in addition to their magnitude to be fully described.

🎯 Exam Tip: Understand that if acceleration has a direction (e.g., speeding up in a certain direction), the force causing it must also have a direction (Newton's Second Law, \( F=ma \)). This concept is fundamental to understanding motion.

Based on the explicit directive to process and map ONLY the questions located between page 15 and page 16 of the provided OCR content, there are no questions or answers found within this specific page range. The content on page 15 consists of a "Search" bar and a list of "Recent Posts" (which are navigation links, not questions), and page 16 contains only copyright information and a final link.