A body of mass 2kg is resting on a rough horizontal surface. A force of 20N is now applied to it for 10s, parallel to the surface. If the coefficient of kinetic friction between the surfaces of contact is 0.2. Calculate
change in kinetic energy of the object in 10s. (Take `g=10ms^(-2)`).

Explanation

Mass = 2kg
Initial velocity
`u=0`
Force `F=2N`
Coefficient of kinetic friction `mu_(k)=0.2`
Time `t=10s`
`f_(k)=mu_(k)mg`
`=0.2xx2xx10=4N`
Net force `F'=f-f_(k)`
`=20-4=16N`
Acceleration,
`a=(F')/(m)=(16)/(2)=8ms^(-1)`
Distance covered
`s=ut+(1)/(2)at^(2)`
`s=0+(1)/(2)xx8xx(10)^(2)=400m`
Change in kinetic energy
= Work done by the net force
`W_(2)=F's`
`=16xx400=6400J`