**Question:**A $2$ kg body is dropped from height of $1$ m onto a spring of spring constant $800$ N/m as shown in the figure. A frictional force equivalent to $0.4$ kgwt acts on the body. What will be the speed of the body just before striking the spring? (Take $g=10m/s_{2}$)

**Solution:**We know that the change of kinetic energy is equal to work done by the system.

Change of Kinetic energy $=21 mv_{2}−0=21 mv_{2}$

Work done $=(mg−f)h$ , where $f=0.4kgwt=0.4×10=4N$, friction force.

Now, $21 mv_{2}=(mg−f)h$

$21 2v_{2}=(20−4)1$

$v_{2}=16$

$v=4m/s$