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A block of mass 1 kg is placed on a truck which accelerates with acceleration 5 m/s². The coefficient of static friction between the block and truck is 0.6. The frictional force acting on the block is:

1. 5 N
2. 6 N
3. 5.88 N
4. 4.6 N

Correct answer: 5 N

Solution

The maximum static friction force is given by \( f_s = \mu_s \cdot m \cdot g \), where \( \mu_s = 0.6 \), \( m = 1 \) kg, and \( g = 9.8 \, \text{m/s}^2 \). This gives \( f_s = 0.6 \cdot 1 \cdot 9.8 = 5.88 \, \text{N} \). However, the required frictional force to prevent slipping is \( f = m \cdot a = 1 \cdot 5 = 5 \, \text{N} \), which is less than the maximum static friction. Thus, the frictional force acting is 5 N.

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