A block m 1 rests on a frictionless surface. A second block m 2 sits on top of t

Question

A block m1 rests on a frictionless surface. A second block m2 sits on top of the first block. Here m1 = m2 = m. A horizontal force F is applied to the bottom block to pull it to the right as shown below. Assume that the blocks move together across the surface (block m2 does not slide off of block m1).

(a) What is the magnitude of the acceleration of the blocks? (Use the following as necessary: m, g, and F.)
a =

1

(b) What are the magnitude and direction of the static friction force acting on m1 due to m2? (Use the following as necessary: m, g, and F.)

2

(c) What are the magnitude and direction of the static friction force acting on m2 due to m1? (Use the following as necessary: m, g, and F.)

4

2

Explanation / Answer

es we can apply newton's second law

total mass = 2m

acceleration = F/2m

b)

friction always opposes relative motion

Ans: leftwards

magnitude = m2*a = F/2

c)

magnitude is

F/2 and direction is rightwards [this provides necessary acceleration for m2 ]

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