Question

a body of mass 10kg is dropped from height of 20 m (a) finds its potential energy before it dropped (b) find its kinetic energy when it is 8 m above the ground​

Answer 1

Answer:

At any point of the time total sum of P.E + K.E will always be constant

so, P.E at start = mgh = 10*10*20 = 2000J

and loss in P.E when it is at 8 m above the ground will be equal to gain in K.E

so, K.E = mg(20-8) = 10*10*12 = 1200J

and K.E at bottom = P.E at top = 2000J

Answer 2

Answer:

P.E is = mgh

Explanation:

m=10kg
g=9.8
h=20m

10*9.8*20=1960j