Advertisements
Advertisements
प्रश्न
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.
Advertisements
उत्तर
Given load m = 200kgf, displacement h = 25 m, time = 5s, g = 10 N kg-1
Now work done = 200 kgf x 25 = 50000 J
or power = `"work done"/"time taken" = 5000/5` = 1000 watt
APPEARS IN
संबंधित प्रश्न
Can there be displacement of an object in the absence of any force acting on it? Think. Discuss this question with your friends and teacher.
Derive a relationship between S.I. and C.C.S. unit of work
Define work.
The work done on an object does not depend on the :
Draw a neat labeled diagram to show the direction of two forces acting on a body to produce rotation in it. Also mark the point about which rotation takes place.
A boy does work when he pushes against a brick wall. (yes/no).
Is work done a scalar or a vector physical quantity?
How many joules are there in 1 mega joule?
Fill in the boxe to show the corresponding energy transformation.
A car is accelerated on a levelled road and attains a velocity 4 times of its initial velocity. In this process the potential energy of the car
