In the gure below, a slab of mass m = 40 kg rests on a frictionless floor, and a

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Question

In the gure below, a slab of mass m = 40 kg rests on a frictionless floor, and a block ofrmass m2 10 kg rests on top ofthe slab. Between block and slab the co effaent of static friction is and the coefficient of kinetic friction is 0.33. The block is pulled by a horizontal force with a magnitude of 100 N. (Assume that the +x axis is to the right.) 50 m2 =0 (a) What is the resulting acceleration of the block? (Express your answer in vector form.) ablock What is the normal force on the block? What is the maximum possible static friction on the block? Does the applied force exceed that value? What is the actual frictional force on the block? What, then, is the frictional force on the slab? Is there any other horizontal force on the slab? What, then, is the acceleration of the slab? m/s2 (b) What is the resulting acceleration of the slab? (Express your answer in vector form.) aslab What is the normal force on the block? What is the maximum possible static friction on the block? Does the applied force exceed that value? What is the actual frictional force on the block? What, then, is the frictional force on the slab? Is there any other horizontal force on the slab? What, then, is the acceleration of the slab? m/s2 GO Tutorial Additional Materials Section 6.1

Explanation / Answer

Limiting friction between block and slab f = usmg =.5×10×9.8

f = 49 N

Now, let assume both move together

So, by fbd of system acceleration

a = 100/(10+40) = 2 m/s^2

Now, by fbd of slab

f = m2a = 40×2 =80 N. This is not possible since limiting friction is 49 N

So block slip on slab and not moving together

Now kinetic friction acts between them

fk = ukm1g = .33×10×9.8 = 32.34 N

Now, by fbd of block

100- 32.34 = 10 a2

a2 = 6.77 m/s^2

Now by fbd of slab

a1 = 32.34/40 = .81m/s^2

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In the gure below, a slab of mass m = 40 kg rests on a frictionless floor, and a

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