TG EAMCET 3 May 2025 Shift 2 Paper

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