A block of mass 2.52 kg is placed on top of a second block of mass 5.01 kg. The

Question

A block of mass 2.52 kg is placed on top of a second block of mass 5.01 kg. The coefficient of kinetic friction between the 5.01 kg block and the surface is 0.23. A horizontal force F is applied to the 5.01 kg block.

2.52kg
5.01 kg
µb = 0.23

Calculate the magnitude of the force neces- sary to pull both blocks to the right with an acceleration of 2.57 m/s2. Assume no slipping between the two blocks. The acceleration of gravity is 9.8 m/s2 .
Answer in units of N

Find the minimum coefficient of static friction µt between the blocks such that the 2.52 kg block does not slip under an acceleration of 2.57 m/s2.

Explanation / Answer

force necessary to pull both blocks = .23*7.53*9.8 + 7.53*2.57= 36.3N minimum coefiicient of static friction = 2.57/9.8 = 0.26

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