A body initially at rest is acted upon by a constantforce. The rate of

1.	A body initially at rest is acted upon by a constantforce. The rate of change of its kinetic energyvaries
Answer» Basically the request is for deriving the expression for Kinetic Energy. It is given that mass 'm' at rest whose velocity changes to 'v' in 't' seconds . It means there is acceleration and hence a force F is acting on the body. First we will compute the acceleration = a from the relation V= U+at In our case U the initial velocity = 0 ∴ V = at ⇒ a = V/t Force = ma = mV/t Distance moved is (1/2)at² = (1/2)(V/t)t² = Vt/2 Work done on the body : (which is stored as Kinetic energy in it) = KE = Fs = (ma)s = m(V/t)* (Vt/2) = (1/2)mV²

A body initially at rest is acted upon by a constantforce. The rate of change of its kinetic energyvaries

Answer»

Basically the request is for deriving the expression for Kinetic Energy.

It is given that mass 'm' at rest whose velocity changes to 'v' in 't' seconds .

It means there is acceleration and hence a force F is acting on the body.

First we will compute the acceleration = a from the relation V= U+at

In our case U the initial velocity = 0 ∴ V = at ⇒ a = V/t

Force = m*a = m*V/t

Distance moved is (1/2)*a*t² = (1/2)*(V/t)*t² = Vt/2

Work done on the body :

(which is stored as Kinetic energy in it) = KE = F*s = (m*a)*s = m*(V/t)* (Vt/2)

= (1/2)*m*V²