Advertisements
Advertisements
Question
A water pump raises 50 kg of water through a height of 25 m in 5s. Calculate the power supplied by the pump (Take: g = 10 N kg−1).
Advertisements
Solution
Given m = 50 kg, g = 10N kg−1, d = 25 m, t = 5s.
Force of gravity F = mg
= 50 × 10
= 500 N
Work done W = F × d
= 500 × 25
= 12,500 J
Power P =`"W"/"t" = (12,500)/5=2500"W".`
APPEARS IN
RELATED QUESTIONS
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.
What should be the angle between the direction of force and the direction of motion of a body so that the work done is zero?
A weight lifted a load of 200 kgf to a height of 2.5 m in 5 s. calculate: (i) the work done, and (ii) the power developed by him. Take g= 10 N kg-1.
A boy pulls a box up to 10 m with a force of 5 kgf. Calculate the work done by him.
State the energy changes which take place when A toy car with a wound spring moves on ground.
Calculate the work done when:
A 5 kg weight is lifted 10 m vertically. (g = 9.8 m/s2)
A girl of mass 40 kg runs a height of 80 stairs, each 25 cm high with a load of 20 kg on her head in 25 sec. If g is 10 m/s2, find:
(i) Gravitational force acting on the girl.
(ii) Work done by her.
(iii) Useful work done by her.
(iv) Her power in watt.
Force acting on a particle moving in a straight line varies with the velocity of the particle v as F = `K/v`, where K is a constant. The work done by this force in time t is
A body of mass 0.5 kg travels on straight line path with velocity v = (3x2 + 4) m/s. The net work done by the force during its displacement from x = 0 tox = 2 m is ______.
A force of F = (5y + 20) `hat"j"` N acts on a particle. The work done by this force when the particle is moved from y = 0 m to y = 10 m is ______ J.
