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A 42-kg slab rests on a frictionless floor. A 9.7-kg block rests on top of the slab, as in Fig. 5-40. The coefficient of static friction between the block and the slab is 0.53, while the co- efficient of kinetic friction is 0.38. The 9.7-kg block is acted on by a horizontal force of 110 N. What are the resulting accelerations of (a) the block and (b) the slab?

A 42-kg slab rests on a frictionless floor. A 9.7-kg block rests on top of the slab, as in Fig. 5-40. The coefficient of static friction between the block and the slab is 0.53, while the co- efficient of kinetic friction is 0.38. The 9.7-kg block is acted on by a horizontal force of 110 N. What are the resulting accelerations of (a) the block and (b) the slab?

Grade:upto college level

1 Answers

Navjyot Kalra
askIITians Faculty 654 Points
8 years ago

232-541_1.PNG
The force F acting on the block has a magnitude of 110 N, and is greater than the static force of friction. Therefore the block slides on one another.
(a)
Apply Newton’s second law of motion to block,

232-765_1.PNG
= 7.61 m/s2
Round off to two significant figures,
232-2065_1.PNG
(b)
Apply Newton’s second law of motion to slab,

232-391_1.PNG
Therefore, the acceleration of the slab is 0.86 m/s2

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