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

Question

A wood block of mass M1 =10 kg rests on a concrete slab of mass M2 = 40 kg The slab tests on a frictionless 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 f , is a variable force 11k Coefficient only gives you its maximum possible value.) Case I: P = 80 N. Case 2: P = 40

Explanation / Answer

Do your FBD and m1 should accelerate to the left with this equation because it will overcome the static friction:

Static friction check:

Force needed to overcome static friction

s*m1(g) = 58.86

Force of case 1 = 80N

So we use kinetic:

80 - k*m1(g) = m1a

80 - 39 = ma

41 = m1a

41/m1 = 4.1m/s2 to the left

 

Now B should feel the friction force of m1 because its a newton-pair:

B should feel the force generated by the push which is:

k*m1(g)

But it is equal now to the combined mass of m1+m2

k*m1(g) = (m1+m2)a

39.24 = (40+10)a

a = 39.24/50

a = .7848 to the left

 

2) For case two, the force 40N Will not overcome the force of static friction required (58.86)

so a of m1 = 0

now a of m2 is just

F=ma

where m is m1+m2

40 = (10+40)a

40/50 = a = 0.8m/s^2   to the left

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