A wood block of mass M1 =10 kg rests on a concrete slab of mass M2 =40 kg The sl

Question

A wood block of mass M1 =10 kg rests on a concrete slab of mass M2 =40 kg The slab rests on a functionless surface. The coefficient of static friction between the slab and block is 0.6. The coefficient of kinetic friction is 0.4. A horizontal force P is applied to the block. For both cases below, find the horizontal acceleration of each mass and the force of friction (0 between them. (Careful! Remember that fstatic, is a variable force 11k Coefficient only gives you its maximum possible value.) Case I: P = 80 N. Case 2: P= 40 N.

Explanation / Answer

(a)

Normal=98 N

Maximum static friction=98*0.6=58.8 N

P=80 N

Thus,P is grater than friction.

Kinetic friction=98*0.4=39.2 N

A1=80-39.2/10

=4.08 m/s^2

A2=39.2/40=0.98 m/s^2

(b)

P=40 N

Maximum static friction=98*0.6=58.8 N

So, above block will remain attach with the lower block.

Both will move with same acceleration = 40/50

=0.8 m/s^2

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