Question

A ball of mass 10 kg is located at the top of the hill(point A). The total height of the hill from ground is 100m. The ball started falling down What will be its kinetic energy when the ball
will be at height of 50 m (point B) from ground? Show that the total energy ( kinetic + potential) remains conserved in this journey from point A to point B.(ignore air resistance,
g=10m/s^2)
pls give aswer fast ​

Answer 1

Answer:

Given,

Mass (m) = 10 kg

Height (h) = 100 m

Acceleration due to gravity (g) = 9.8 m/s²

i. Taking h as 50 m,

Initial Velocity (u) = 0 m/s

v² = u²+2gh

v² = 2*9.8*50

v² = 980

K.E = 1/2mv²

= 1/2 * 10 * 980

= 4900 J

ii. K.E at Point A = 1/2*m*v²

= 1/2*10*0

= 0 J

P.E at Point A = mgh

= 10*10*100 J

= 10000 J

Total energy at Point A = (0+10000) J

= 10000 J

K.E at Point B = 1/2*m*v²

= 1/2*10*[2gh]

= 1/2*10*2*10*100 J

= 10000 J

P.E at Point B = mgh

= 10*10*0 J

= 0 J

Total Energy at Point B = (10000+0) J

= 10000 J

So,

Total Energy at Point A = Total Energy at Point B

Hence, Proved that energy conserved.

I hope this helps.