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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?

5 years ago

Answers : (1)

Navjyot Kalra
askIITians Faculty
654 Points
							
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
5 years ago
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Course Features

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  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Previous Year Exam Questions


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