Question

the same body is Now taken from a height of 5m to a height of 10 metre find the work done? (g=10m s-2)​

Answer 1

Answer:

Work done is change in potential energy of the object. Here the assumption is that the force which lifts the body, balances the weight at all times, meaning which there is no linear velocity of the object during this motion.

All the work done by the force is reflected as the increase in the stored potential energy of the object. Potential energy is defined as a difference of potential energy between two reference points and

U2 - U1 = mgh = 10 * 10 * 5 = 500 N-m

So the work done =Change in potential energy= 500 Nm.

Explanation:

$Hope \: it \: helps :)$

Answer 2

Explanation:

m=5kg,h1=5m,h2=10m,g=10ms−2

Increase in potential energy =(final potential energy) − (initial potential energy) = mg×(h2−h1)=5×10×(10−5)=250J