On the horizontal surface of a truck ($\mu = 0.6$), a block of mass 1 kg is placed. If the truck is accelerati

Question

On the horizontal surface of a truck ($\mu = 0.6$), a block of mass 1 kg is placed. If the truck is accelerating at the rate of $5 	ext{ m/s}^2$, then the frictional force on the block will be:

Accepted Answer

1. **Analyze Forces:** The truck accelerates at $a = 5 	ext{ m/s}^2$. For the block of mass $m=1 	ext{ kg}$ to move along with the truck (stay stationary relative to it), a force is required. In the frame of the truck, a pseudo force $F_p = ma$ acts on the block opposite to the acceleration.
   $$ F_{req} = ma = 1 	ext{ kg} 	imes 5 	ext{ m/s}^2 = 5 	ext{ N} $$
2. **Calculate Limiting Friction:** The maximum possible static friction ($f_{max}$) is determined by the coefficient of friction $\mu$ and the normal reaction $N$ ($N=mg$).
   $$ f_{max} = \mu mg = 0.6 	imes 1 	ext{ kg} 	imes 9.8 	ext{ m/s}^2 = 5.88 	ext{ N} $$
   (If $g=10 	ext{ m/s}^2$, $f_{max} = 6 	ext{ N}$).
3. **Conclusion:** The required force ($5 	ext{ N}$) is less than the limiting friction ($5.88 	ext{ N}$). Therefore, the block does not slide. Since static friction is **self-adjusting** (it adjusts to match the applied force up to the limit), the frictional force acting on the block is exactly equal to the force required to accelerate it [Source 67, 68].
   $$ f_s = F_{req} = 5 	ext{ N} $$

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