A 2 kg block on a horizontal surface experiences a static friction μs = 0.4. If the impending motion would start at F = 8 N, what is the maximum static friction force?

A 2 kg block on a horizontal surface experiences a static friction μs = 0.4. If the impending motion would start at F = 8 N, what is the maximum static friction force?

• A. 7.84 N
• B. 8.00 N
• C. 0 N
• D. 9.8 N

Answer: (A)

Static friction can supply up to a maximum value equal to μs times the normal force. On a horizontal surface the normal force is just the weight, N = mg. With m = 2 kg and g ≈ 9.8 m/s^2, N = 2 × 9.8 = 19.6 N. The maximum static friction is f_s,max = μs N = 0.4 × 19.6 = 7.84 N. This is the largest opposing friction force the surface can provide while the block remains at rest. If you tried to push with 8 N, that would exceed the maximum static friction, so motion would begin. The given threshold of 8 N is not consistent with the numbers, the actual threshold here is 7.84 N.

Static friction can supply up to a maximum value equal to μs times the normal force. On a horizontal surface the normal force is just the weight, N = mg. With m = 2 kg and g ≈ 9.8 m/s^2, N = 2 × 9.8 = 19.6 N. The maximum static friction is f_s,max = μs N = 0.4 × 19.6 = 7.84 N. This is the largest opposing friction force the surface can provide while the block remains at rest. If you tried to push with 8 N, that would exceed the maximum static friction, so motion would begin. The given threshold of 8 N is not consistent with the numbers, the actual threshold here is 7.84 N.