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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?

Amit Saxena

11 Years agoGrade upto college level

1 Answer

Navjyot Kalra

11 Years ago

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,

= 7.61 m/s²
Round off to two significant figures,

(b)
Apply Newton’s second law of motion to slab,

Therefore, the acceleration of the slab is 0.86 m/s²

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