CBSE Class 11 Physics Work Energy Power Advanced Problems

Class 12 Physics: Work, Energy and Power (Advanced Level Assignment)

A 0.5 kg block slides from point A on a horizontal track with an initial speed of 3 m/s towards a weightless spring of length 1 m and having a force constant 2 N/m. The part AB of the track is frictionless and the part BC has coefficient of static and kinetic friction as 0.22 and 0.20 respectively. If the distances AB and BD are 2 m and 2.14 m respectively, find the total distance through which the block moves before it comes to rest completely. (\(g=10\ \text{m/s}^2\))

Answer: 4.24 m

A particle is suspended from a fixed point by a string of length 5 m. It is projected horizontally from the equilibrium position with such a speed that the string slackens after the particle has reached a height of 8 m above the lowest point. Find the speed of the particle just before the string slackens and the height to which the particle will rise further.

Answer: Speed just before the string slackens \(=5.42\ \text{m/s}\) (taking \(g=9.8\ \text{m/s}^2\)); it rises a further \(0.96\ \text{m}\).

Note: the printed key gives 0.97 m; working it out exactly gives \(v^2=3g=29.4\), \(h=\dfrac{(0.8v)^2}{2g}=0.96\) m.

One end of a light spring of natural length \(d\) and spring constant \(k\) is fixed on a rigid wall and the other end is fixed to a smooth ring of mass \(m\) which can slide without friction on a vertical rod fixed at a distance \(d\) from the wall. Initially the spring makes an angle of 37° with the horizontal as shown in the figure. If the system is released from rest, find the speed of the ring when the spring becomes horizontal. [\(\sin37^\circ=3/5\)]

Answer: \[ v=d\sqrt{\frac{3g}{2d}+\frac{k}{16m}} \]

A ring of mass \(m=0.3\) kg slides over a smooth vertical rod A. Attached to the ring is a light string passing over a smooth fixed pulley at a distance of 0.8 m from the rod as shown in the figure. At the other end of the string there is a mass \(M=0.5\) kg. The ring is held in level with the pulley and then released.

1. Determine the distance by which the mass \(m\) moves down before coming to rest for the first time.
2. How far below the initial position of \(m\) is the equilibrium position of \(m\) located?

Answer: (a) 1.5 m    (b) 0.6 m

A string, with one end fixed on a rigid wall, passing over a fixed frictionless pulley at a distance of 2 m from the wall, has a point mass \(M=2\) kg attached to it at a distance of 1 m from the wall. A mass \(m=0.5\) kg attached at the free end is held at rest so that the string is horizontal between the wall and the pulley and vertical beyond the pulley. Find the speed with which the mass \(M\) will hit the wall when the mass \(m\) is released.

Answer: \(v\approx3.4\ \text{m/s}\) (the key gives 3.3 m/s).

Note: \(M\) swings on a circle of radius 1 m about the wall point and reaches the wall directly below it, where the string to the pulley has length \(\sqrt5\) m, so \(m\) is lifted by \(\sqrt5-1\) m and \(u_m=\dfrac{2}{\sqrt5}v_M\). Energy: \(2g(1)-0.5g(\sqrt5-1)=\tfrac12(2)v^2+\tfrac12(0.5)\tfrac45v^2\), which gives \(v=3.36\) m/s for \(g=9.8\ \text{m/s}^2\) (3.39 m/s for \(g=10\)). The key value 3.3 m/s is this number cut short.

Two blocks \(A\) and \(B\) each having mass 0.32 kg are connected by a light string passing over a smooth pulley as shown in the figure. The horizontal surface on which the block \(A\) slides is smooth. The block \(A\) is attached to a spring of force constant 40 N/m whose other end is fixed to a support 0.40 m above the horizontal surface. Initially, when the system is released to move, the spring is vertical and unstretched. Find the velocity of the block \(A\) at the instant it breaks off the surface below it. [\(g=10\ \text{m/s}^2\)]

Answer: 1.54 m/s

A block of mass \(m\) is held at rest on a smooth horizontal floor. A light frictionless small pulley is fixed at a height of 6 m from the floor. A light inextensible string of length 16 m, connected with \(A\), passes over the pulley and another identical block \(B\) is hung from the string. Initial height of \(B\) is 5 m from the floor as shown in the figure. When the system is released from rest, \(B\) starts to move vertically downwards and \(A\) slides on the floor towards the right.

1. If at an instant the string makes an angle \(\theta\) with the horizontal, calculate the relation between velocity \(u\) of \(A\) and velocity \(v\) of \(B\).
2. Calculate \(v\) when \(B\) strikes the floor. (\(g=10\ \text{m/s}^2\))

Answer: (a) \(v=u\cos\theta\)    (b) \( v=\dfrac{40}{\sqrt{41}}\ \text{m/s}\approx6.25\ \text{m/s} \)

A horizontal plane supports a plank with a bar of mass 1 kg placed on it and attached by a light elastic non-deformed cord of length 40 cm to a point O as shown in the figure. The coefficient of friction between the bar and plank is 0.2. The plank is slowly shifted to the right until the bar starts sliding over it. It occurs at the moment when the cord deviates from the vertical by 30°. Find the work that has been performed till that moment by the friction force acting on the bar in the reference frame fixed to the plane.

Answer: 0.09 J

A body of mass \(m\) was slowly pulled up the hill as shown in the figure, by a force \(F\) which at each point was directed along a tangent to the trajectory. Find the work performed by this force, if the height of the hill is \(h\), the length of its base \(l\), and the coefficient of friction between \(m\) and the hill is \(\mu\).

Answer: \( W=mgh+\mu mgl \)

A chain \(AB\) of length \(l\) is loaded in a smooth horizontal tube so that a part of its length \(h\) hangs freely and touches the surface of the table with its end \(B\). At a certain moment, the end \(A\) of the chain is set free. With what velocity will this end of the chain slip out of the tube?

Answer: \[ v=\sqrt{2gh\,\ln\frac lh} \]

A system consists of two identical cubes, each of mass \(m\), linked together by a compressed weightless spring of spring constant \(k\). The cubes are also connected by a thread which is burned through at a certain moment. Find:

1. at what values of \(\Delta l\), the initial compression of the spring, the lower cube will bounce up after the thread has been burned through;
2. to what height \(h\) the centre of gravity of this system will rise if the initial compression of the spring is \(\Delta l=7mg/k\).

Answer: (a) \( \Delta l>\dfrac{3mg}{k} \)    (b) \( h=\dfrac{8mg}{k} \)

A nail is located at a certain distance vertically below the point of suspension of a simple pendulum. The pendulum bob is released from the position where the string makes an angle of 60° with the downward vertical. Find the distance of the nail from the point of suspension such that the bob will just perform a complete revolution with the nail as centre. The length of the pendulum is 1 m.

Answer: 0.8 m

A particle is suspended vertically from a point \(O\) by an inextensible massless string of length \(L\). A vertical line \(AB\) is at a distance of \(L/8\) from \(O\) as shown in the figure. The particle is given a horizontal velocity \(u\). At some point, its motion ceases to be circular and eventually the object passes through the line \(AB\). At the instant of crossing \(AB\), its velocity is horizontal. Find \(u\).

Answer: \( u=2.14\sqrt{gL} \)

A spherical ball of mass \(m\) is kept at the highest point in the space between two fixed, concentric spheres \(A\) and \(B\) (see figure). The smaller sphere \(A\) has a radius \(R\) and the space between the two spheres has a width \(d\). The ball has a diameter very slightly less than \(d\). All surfaces are frictionless. The ball is given a gentle push (towards the right in the figure). The angle made by the radius vector of the ball with the upwards vertical is denoted by \(\theta\) (shown in the figure).

1. Express the total normal reaction force exerted by the spheres on the ball as a function of angle \(\theta\).
2. Let \(N_A\) and \(N_B\) denote the magnitudes of the normal reaction forces on the ball exerted by the spheres \(A\) and \(B\), respectively. Sketch the variations of \(N_A\) and \(N_B\) as functions of \(\cos\theta\) in the range \(0\le\theta\le\pi\) by drawing two separate graphs in your answer book, taking \(\cos\theta\) on the horizontal axes.

Answer: (a) \( N=mg(3\cos\theta-2) \) (taken positive when directed away from \(O\), i.e. exerted by sphere \(A\); negative means directed towards \(O\), exerted by sphere \(B\)).

(b) For \( \theta\le\cos^{-1}(2/3) \): \( N_B=0,\ N_A=mg(3\cos\theta-2) \). For \( \theta\ge\cos^{-1}(2/3) \): \( N_A=0,\ N_B=mg(2-3\cos\theta) \).

A smooth light horizontal rod \(AB\) can rotate about a vertical axis passing through its end \(A\). The rod is fitted with a small sleeve of mass \(m\) attached to the end \(A\) by a weightless spring of length \(l_0\) and spring constant \(k\). What work must be performed to slowly get this system going and reaching the angular velocity \(\omega\)?

Answer: \[ W=\frac{k\,l_0^{2}\,x(1+x)}{2(1-x)^{2}},\qquad x=\frac{m\omega^{2}}{k} \]

A small block of mass \(m\) slides along a frictionless loop-inside-loop track. What should be the ratio of the radius of the outer loop to the radius of the inner loop so that the block may not lose contact at the highest point of the inner loop?

Answer: \( R:r=5:2 \)

Note: the source PDF gives no figure for this question, so the starting condition is not known and this printed answer could not be checked independently.

Figure shows a smooth track which consists of a straight inclined part of length \(L\) joining smoothly with the circular part. A particle of mass \(m\) is projected up the incline from its bottom.

1. Find the minimum projection speed \(u_0\) for which the particle reaches the top of the track.
2. Assuming that the projection speed is \(2u_0\) and that the block does not lose contact with the track before reaching its top, find the force acting on it when it reaches the top.
3. Assuming that the projection speed is only slightly greater than \(u_0\), where will the block lose contact with the track?

Answer: (a) \( u_0=\sqrt{2g\left[R(1-\cos\theta)+L\sin\theta\right]} \)

(b) \( 6mg\left(1-\cos\theta+\dfrac LR\sin\theta\right) \) (this is \(mv^2/R\), the net force towards the centre at the top, since \(v^2=6g\left[R(1-\cos\theta)+L\sin\theta\right]\)).

(c) The radius through the particle makes an angle \( \cos^{-1}\!\left(\dfrac23\right) \) with the vertical.

A chain of mass \(m\) and length \(l\) is placed symmetrically on a pulley of mass \(M\) (\(=2m\)) and radius \(R\) (\(=l/8\)) as shown. If the chain is given a slight pull on one side, find the velocity of the chain when the other end reaches the highest position on the pulley. The surface of the pulley in contact with the chain is rough enough to prevent any slipping.

Answer: \[ V=\sqrt{\frac{136-\pi^{2}}{512}\,g\,l} \]

Note: this answer holds when the pulley is treated as a uniform disc (moment of inertia \(\tfrac12MR^2\)); the question does not state this.

A body of mass \(m\) is hauled from the Earth surface by applying a force \(F\) varying with the height of ascent \(y\) as \(F=2(1-ay)\,mg\), where \(a\) is a positive constant. Find

1. the work performed by this force;
2. the increment of the body potential energy in the gravitational field of the Earth over the first half of the ascent.

Answer: (a) \( \dfrac{3mg}{4a} \)    (b) \( \Delta U=\dfrac{mg}{2a} \)

Note: the source prints the force as \(2(ay-1)mg\), which would push the body downwards at the start. The printed answers match \(F=2(1-ay)mg\), so this form is used. The whole ascent ends where the speed is zero: \(\int_0^H2(1-ay)mg\,dy=mgH\) gives \(H=1/a\); the first half is \(y=0\) to \(1/(2a)\).

The system of masses \(A\) and \(B\) shown in the figure is released from rest with \(x=0\). Determine

1. the velocity of mass \(B\) when \(x=3\) m;
2. the maximum displacement of mass \(B\).

Answer: (a) \( v\approx3.96\ \text{m/s}\approx4\ \text{m/s} \) (taking \(g=10\ \text{m/s}^2\)); the printed key gives 5 m/s.

(b) \( 8\sqrt2\ \text{m}\approx11.3\ \text{m} \)

Note: at \(x=3\) m each string segment is \(\sqrt{16+9}=5\) m, so \(A\) has risen \(2(5)-8=2\) m and \(v_A=\dfrac{2x}{\sqrt{16+x^2}}v_B=1.2\,v_B\). Energy: \(mg(3)-\dfrac{m}{\sqrt2}g(2)=\tfrac12mv_B^2+\tfrac12\dfrac m{\sqrt2}(1.2v_B)^2\), so \(v_B^2=15.7\ \text{m}^2/\text{s}^2\). The same geometry gives the key answer \(8\sqrt2\) m for part (b), and the speed of \(B\) cannot reach 5 m/s anywhere in its motion, so the printed 5 m/s looks like a key error.