Advertisements
Advertisements
प्रश्न
A block of mass 20 kg is pulled up a slope (fig.12) with a constant speed by applying a force of 500 N parallel to the slope. A and B are initial and final positions of the block.
(a) Calculate the work done by the force in moving the block from A and B.
(b) Calculate the potential energy gained by the block.
Advertisements
उत्तर
(a) given force = 500N
Vertical displacement = Final position - initial position = BC = 4m
work done = force x displacement = 500 x 4 = 2000J
(b) P.E gained = mgh = (mg) h = 500 x 4 = 2000J
APPEARS IN
संबंधित प्रश्न
Work is said to be done by a forte only when ................
Work done = Force × ........................................
A body, when acted upon by a force of 10 kgf, moves to a distance 0.5 m in the direction of force. Find the work done by the force. Take 1 kgf = 10 N.
Express joule in terms of erg.
It takes 20 s for a person A of mass 50 kg to climb up the stairs, while another person B does the same in 15 s. Compare the work done.
A vessel containing 50 kg of water is placed at a height 15 m above the ground. Assuming the gravitational potential energy at ground to be zero, what will be the gravitational potential energy of water in the vessel? (g = 10 m s-2)
Determine the amount of work done when an object is displaced at an angle of 30° with respect to the direction of the applied force.
A uniform circular motion is an accelerated motion. Justify it.
A proton is kept at rest. A positively charged particle is released from rest at a distance d in its field. Consider two experiments; one in which the charged particle is also a proton and in another, a positron. In the same time t, the work done on the two moving charged particles is ______.
