Total mass of truck and block M′=8+2‌=10ton=10×1000kg ‌=104kg Acceleration produced due to application of 25 kN force, a=‌
F
M′
=‌
25×103
104
=2.5m∕s2 Maximum frictional force acting on the block Fsm‌=µsmg ‌=0.3×2000×10=6000N The force required to decelerate the block with the same acceleration as the truck, F′=M‌block ‌×a=2000×2.5=5000N acceleration. Thus, frictional force acting on the block =5000N
