Question

A body of mass 500 Kg is dropped from a height of 10 m from the ground. Calculate the

potential energy at a height of 7 m from the ground .Also calculate the kinetic energy when

it reaches the ground. [ g=10m/s2

]​

Answer 1

Answer:

Solution :

Here mass of the ball, `m =5kg`

height of the ball, `h= 10m`

(a) initial potential energy of the ball,

`E_p = mgh = (5kg)(10m//s^2)(10m) = 500J`

(b) When the ball reaches the ground, its potential energy becomes zero as it is entirely converted into

its kinetic energy `(E_k),i.e.,`

`E_k =500J ...(i)`

(c ) It `upsilon` is the velocity attained by the ball before reaching the ground,

`E_k =(1)/(2)m upsilon^2`

or `upsilon =sqrt((2E_k)/(m)) = sqrt((2xx500J)/(5kg))= 14.14m//s`

Answer 2

Explanation:

Here mass of the ball, m=5kg

Here mass of the ball, m=5kgheight of the ball, h=10m

Here mass of the ball, m=5kgheight of the ball, h=10m(a) initial potential energy of the ball,

Here mass of the ball, m=5kgheight of the ball, h=10m(a) initial potential energy of the ball,Ep=mgh=(5kg)(10m/s2)(10m)=500J

Here mass of the ball, m=5kgheight of the ball, h=10m(a) initial potential energy of the ball,Ep=mgh=(5kg)(10m/s2)(10m)=500J(b) When the ball reaches the ground, its potential energy becomes zero as it is entirely converted into

Here mass of the ball, m=5kgheight of the ball, h=10m(a) initial potential energy of the ball,Ep=mgh=(5kg)(10m/s2)(10m)=500J(b) When the ball reaches the ground, its potential energy becomes zero as it is entirely converted intoits kinetic energy (Ek),i.e.,

Here mass of the ball, m=5kgheight of the ball, h=10m(a) initial potential energy of the ball,Ep=mgh=(5kg)(10m/s2)(10m)=500J(b) When the ball reaches the ground, its potential energy becomes zero as it is entirely converted intoits kinetic energy (Ek),i.e.,Ek=500J...(i)

Here mass of the ball, m=5kgheight of the ball, h=10m(a) initial potential energy of the ball,Ep=mgh=(5kg)(10m/s2)(10m)=500J(b) When the ball reaches the ground, its potential energy becomes zero as it is entirely converted intoits kinetic energy (Ek),i.e.,Ek=500J...(i)(c ) It υ is the velocity attained by the ball before reaching the ground,

Here mass of the ball, m=5kgheight of the ball, h=10m(a) initial potential energy of the ball,Ep=mgh=(5kg)(10m/s2)(10m)=500J(b) When the ball reaches the ground, its potential energy becomes zero as it is entirely converted intoits kinetic energy (Ek),i.e.,Ek=500J...(i)(c ) It υ is the velocity attained by the ball before reaching the ground,Ek=12mυ2

Here mass of the ball, m=5kgheight of the ball, h=10m(a) initial potential energy of the ball,Ep=mgh=(5kg)(10m/s2)(10m)=500J(b) When the ball reaches the ground, its potential energy becomes zero as it is entirely converted intoits kinetic energy (Ek),i.e.,Ek=500J...(i)(c ) It υ is the velocity attained by the ball before reaching the ground,Ek=12mυ2or υ=2Ekm−−−−√=2×500J5kg−−−−−−−−√=14.14m/s