A block (4.00-kg) is at rest on the right edge of a large 11.0-kg slab. There is

Question

A block (4.00-kg) is at rest on the right edge of a large 11.0-kg slab. There is no friction below the slab and the horizontal surface it is on but there is friction between the block and the slab. The coefficient of kinetic friction between the two is 0.25. The slab is 3.00 meters wide and a constant horizontal force is applied to the left on the block. If it takes 2.00 s for the block to reach the other side of the slab,

(a) what is the magnitude of the force applied to the block?

(b) How far does the bottom block travel?

(c) What is the final velocity of the top and bottom block? (Hint: use a non-moving/accelerating reference frame, i.e. static origin on the right side of the table)

Explanation / Answer

For Block m1

(Normal force acting on the block) N = m1g

balancing the forces, we have

F-F(friction ) =m1a1     (where a1 = acceleration of the mass m1)

F-m1g = m1 a1

a1=F/m1 - g

a1 = F/4 - 0.25*9.8

a1 = F/4 - 2.45

For m2

F(fr) =m2a2

a2 = m1 g/m2

a2 = 0.25*4*9.8* / 11

a2 = 0.89 m/s2

distance x1 = a1t^2 x2 = a2t^2

and x1 = x2+l

a1t^2 = a2t^2 + l

0.5*(F/4 - 2.45) * 2^2 = 0.5*0.89*2^2 + 3

F= 18.97 N

b) x2 = a2t^2 = 0.89*2^2 = 3.56 m

c) final velocities of the block

By using the equation of motion, we have

v=u+at

v1 = 0 + (18.97/4 -2.45) * 2

v1 = 4.585m/s

v2 = 0 + a2t

v2 = 0 + 0.89*2 = 1.78 m/s

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