Advertisements
Advertisements
Question
Consider the situation of the previous question from a frame moving with a speed v0 parallel to the initial velocity of the block. (a) What are the initial and final kinetic energies? (b) What is the work done by the kinetic friction?
Advertisements
Solution
The relative velocity of the ball w.r.t. the moving frame is given by \[v_r = v - v_0\]
(a) Initial kinetic energy of the ball =\[\frac{1}{2}m {v_r}^2 = \frac{1}{2}m(v - v_0 )^2\]
Also, final kinetic energy of the ball =\[\frac{1}{2}m(0 - v_0 )^2 = \frac{1}{2}m {v_0}^2\]
(b) Work done by the kinetic friction = final kinetic energy - initial kinetic energy
APPEARS IN
RELATED QUESTIONS
Is work-energy theorem valid in non-inertial frames?
A ball is given a speed v on a rough horizontal surface. The ball travels through a distance l on the surface and stops. What is the work done by the kinetic friction?
An unruly demonstrator lifts a stone of mass 200 g from the ground and throws it at his opponent. At the time of projection, the stone is 150 cm above the ground and has a speed of 3 m/s. Calculate the work done by the demonstrator during the process. If it takes one second for the demonstrator to lift the stone and throw it, what horsepower does he use?
A block of mass 30 kg is being brought down by a chain. If the block acquires a speed of 40 cm/s in dropping down 2 m, find the work done by the chain during the process.
Consider the situation shown in the following figure. The system is released from rest and the block of mass 1 kg is found to have a speed 0⋅3 m/s after it has descended a distance of 1 m. Find the coefficient of kinetic friction between the block and the table.
A small block of mass 200 g is kept at the top of a frictionless incline which is 10 m long and 3⋅2 m high. How much work was required (a) to lift the block from the ground and put it an the top, (b) to slide the block up the incline? What will be the speed of the block when it reaches the ground if (c) it falls off the incline and drops vertically to the ground (d) it slides down the incline? Take g = 10 m/s2.
A block of mass 5 kg is suspended from the end of a vertical spring which is stretched by 10 cm under the load of the block. The block is given a sharp impulse from below, so that it acquires an upward speed of 2 m/s. How high will it rise? Take g = 10 m/s2.
The bob of a stationary pendulum is given a sharp hit to impart it a horizontal speed of \[\sqrt{3 gl}\] . Find the angle rotated by the string before it becomes slack.
A heavy particle is suspended by a 1⋅5 m long string. It is given a horizontal velocity of \[\sqrt{57} \text{m/s}\] (a) Find the angle made by the string with the upward vertical when it becomes slack. (b) Find the speed of the particle at this instant. (c) Find the maximum height reached by the particle over the point of suspension. Take g = 10 m/s2.
Figure ( following ) 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. Find the minimum projection-speed \[\nu_0\] for which the particle reaches the top of the track.
Figure ( following ) 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.Assuming that the projection-speed is only slightly greater than \[\nu_0\] , where will the block lose contact with the track?
A chain of length l and mass m lies on the surface of a smooth sphere of radius R > l with one end tied to the top of the sphere. Find the gravitational potential energy of the chain with reference level at the centre of the sphere.
A man, of mass m, standing at the bottom of the staircase, of height L climbs it and stands at its top.
- Work done by all forces on man is equal to the rise in potential energy mgL.
- Work done by all forces on man is zero.
- Work done by the gravitational force on man is mgL.
- The reaction force from a step does not do work because the point of application of the force does not move while the force exists.
A bullet of mass m fired at 30° to the horizontal leaves the barrel of the gun with a velocity v. The bullet hits a soft target at a height h above the ground while it is moving downward and emerges out with half the kinetic energy it had before hitting the target.
Which of the following statements are correct in respect of bullet after it emerges out of the target?
- The velocity of the bullet will be reduced to half its initial value.
- The velocity of the bullet will be more than half of its earlier velocity.
- The bullet will continue to move along the same parabolic path.
- The bullet will move in a different parabolic path.
- The bullet will fall vertically downward after hitting the target.
- The internal energy of the particles of the target will increase.
Two bodies of unequal mass are moving in the same direction with equal kinetic energy. The two bodies are brought to rest by applying retarding force of same magnitude. How would the distance moved by them before coming to rest compare?
A raindrop of mass 1.00 g falling from a height of 1 km hits the ground with a speed of 50 ms–1. Calculate
- the loss of P.E. of the drop.
- the gain in K.E. of the drop.
- Is the gain in K.E. equal to a loss of P.E.? If not why.
Take g = 10 ms–2
Suppose the average mass of raindrops is 3.0 × 10–5 kg and their average terminal velocity 9 ms–1. Calculate the energy transferred by rain to each square metre of the surface at a place which receives 100 cm of rain in a year.
