Given,
Mass of bullet is m=0.01kg
Mass of the block is M=4kg
Coefficient=0.25,distance=20m
So, let the speed of the block just after the bullet embedded in it be V and v be the speed of bullet before striking the block,
By applying conservation of momentum,
mv=(m+M)V
V=mvM+m
So, By the work-energy theorem,
12(m+M)V2=Force×displacement=μNd
⇒12(m+M)(mvM+m)2=μ(m+M)gd
mvM+m=2√μgd
V=2√μgd(m+M)m
√0.25×10×20(0.01+4)0.01
=567.09m/s
Thus, the required velocity is =567.09m/s