Advertisements
Advertisements
Question
A weight lifter lifted a load of 200 kgf to a height of 2.5 m in 5 s. Calculate:
- the work done, and
- the power developed by him. Take g = 10 N kg-1
Advertisements
Solution
Force= mg = 200 × 10 = 2000 N
Distance, S = 2.5m
Time , t = 5 s
(i) Work done, W= F × S
W = 2000 × 2.5m = 5000 J
(ii) Power developed = `w/t`= `5000/5` = 1000 W
APPEARS IN
RELATED QUESTIONS
In each of the following a force, F is acting on an object of mass, m. The direction of displacement is from west to east, as shown by the longer arrow. Observe the diagrams carefully and state whether the work done by the force is negative, positive or zero.

1 J = ______ calorie.
Is work a scalar or a vector quantity?
Define 1 joule of work.
Fill in the blank with suitable word :
The work done on a body moving in a circular path is __________
The work done on an object does not depend on the :
A canon ball of mass 500g is fired with a speed of 15m/s-1. Find: its momentum.
A force is applied on a body of mass 20 kg moving with a velocity of 40 ms−1. The body attains a velocity of 50 ms−1 in 2 seconds. Calculate the work done by the body.
