Advertisements
Advertisements
प्रश्न
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.
Advertisements
उत्तर
b and d
Explanation:
When a man of mass m climbs up the staircase of height L, work done by the gravitational force on the man = – mgL
Work done by internal muscular forces = – Work done against gravitational force = mgL
Work done by all the forces = mgL – mgL = 0
As the point of application of the contact forces does not move, hence work done by reaction forces will also be zero.
And work done by friction will also be zero as there a no dissipation or rubbing involved.
APPEARS IN
संबंधित प्रश्न
In figure (i) the man walks 2 m carrying a mass of 15 kg on his hands. In Figure (ii), he walks the same distance pulling the rope behind him. The rope goes over a pulley, and a mass of 15 kg hangs at its other end. In which case is the work done greater?
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 are the initial and final kinetic energies of the ball?
A water pump lifts water from 10 m below the ground. Water is pumped at a rate of 30 kg/minute with negligible velocity. Calculate the minimum horsepower that the engine should have to do this.
In a factory, 2000 kg of metal needs to be lifted by an engine through a distance of 12 m in 1 minute. Find the minimum horsepower of the engine to be used.
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 block of mass 100 g is moved with a speed of 5⋅0 m/s at the highest point in a closed circular tube of radius 10 cm kept in a vertical plane. The cross-section of the tube is such that the block just fits in it. The block makes several oscillations inside the tube and finally stops at the lowest point. Find the work done by the tube on the block during the process.
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 pendulum at rest is given a sharp hit to impart a horizontal velocity \[\sqrt{10 \text{ gl }}\], where l is the length of the pendulum. Find the tension in the string when (a) the string is horizontal, (b) the bob is at its highest point and (c) the string makes an angle of 60° with the upward vertical.
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.
A particle slides on the surface of a fixed smooth sphere starting from the topmost point. Find the angle rotated by the radius through the particle, when it leaves contact with the sphere.
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.
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 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. Suppose the chain is released and slides down the sphere. Find the kinetic energy of the chain, when it has slid through an angle θ.
A smooth sphere of radius R is made to translate in a straight line with a constant acceleration a. A particle kept on the top of the sphere is released at zero velocity with respect to the sphere. Find the speed of the particle with respect to the sphere as a function of the angle θ it slides.
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
A rocket accelerates straight up by ejecting gas downwards. In a small time interval ∆t, it ejects a gas of mass ∆m at a relative speed u. Calculate KE of the entire system at t + ∆t and t and show that the device that ejects gas does work = `(1/2)∆m u^2` in this time interval (neglect gravity).
