A 3.00-kg block rests on the left edge of a 9.00-kg slab, as shown in the figure

Question

A 3.00-kg block rests on the left edge of a 9.00-kg slab, as shown in the figure. Friction is present between the two masses and the coefficient of kinetic friction is 0.25. There is no friction below the slab, between it and the table. There is a constant 12-N force applied horizontally to the top block. If the distance L that the leading edge of the top block travels on the slab is 3.00 m, (a) in what time interval will the top block make it to the right side of the 9.00-kg slab as shown in figure. (b) How far does the 3.00-kg block move in the process? (Hint: use a non-moving/accelerating reference frame, i.e. static origin on the left side of the table)

Explanation / Answer

a)

Normal force from slab to block = mg = 3*9.81 = 29.4 N

Friction force on block = fric coeff*normal force = 0.25*29.4 = 7.36 N

Net force in horizontal direction on block = 12 - 7.36 = 4.64 N

Acceleration of block a = Net force/m = 4.64/3 = 1.55 m/s^2

Using s = ut + 1/2*at^2 we get, 3 = 0 + 1/2*1.55*t^2

Time t = 1.97 s

b)

Force on slab = Friction force = 7.36 N

Acceleration of slab a' = Force/mass = 7.36/9 = 0.818 m/s^2

Again using s = ut + 1/2*at^2 we get

s = 0 + 1/2*0.818*1.97^2 = 1.587 m

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