Two blocks are accelerated across a horizontal frictionless surface as shown. Frictional forces keep the two blocks from sliding relative to each other, and the two move with the same acceleration. If F = 1.2 N and M = 1.0 kg, what is the horizontal component (frictional force) of the force of the small block on the large block?
A) 0.48 N to the right
B) 0.72 N to the right
C) 0.72 N to the left
D) 0.48 N to the left
E) 0.65 N to the left

Question

Two blocks are accelerated across a horizontal frictionless surface as shown. Frictional forces keep the two blocks from sliding relative to each other, and the two move with the same acceleration. If F = 1.2 N and M = 1.0 kg, what is the horizontal component (frictional force) of the force of the small block on the large block? 
A) 0.48 N to the right 
B) 0.72 N to the right 
C) 0.72 N to the left 
D) 0.48 N to the left 
E) 0.65 N to the left

Quizplus · Accepted Answer

First, we must find the acceleration of the system. We can set up an equation using Newton's Second Law for each block:

For the smaller block: 
F - F(friction) = ma

For the larger block: 
F(friction) = Ma

Since the blocks are not sliding relative to each other, we know that the acceleration of both blocks is the same, so we can set the two equations equal to each other:

F - F(friction) = Ma

F(friction) = ma

Substituting the given values, we get:

1.2 N - F(friction) = 2(1.0 kg) a

F(friction) = 1.0 kg a

Solving for a, we get:

a = 0.6 m/s^2

Now we can find the frictional force acting on the smaller block by using the equation:

F(friction) = uN

where u is the coefficient of friction and N is the normal force. Since the blocks are not sliding relative to each other, we know that the normal force acting on the smaller block is equal in magnitude to the weight of the larger block:

N = Mg = (2.0 kg)(9.81 m/s^2) = 19.62 N

Substituting the given values, we get:

F(friction) = uN 
F(friction) = u(19.62 N)

Now we need to find the coefficient of friction, which can be done using the equation:

F(friction) = uN = umg

u = F(friction)/mg

Substituting the given values, we get:

u = (1.0 kg)(0.6 m/s^2)/(2.0 kg)(9.81 m/s^2)

u = 0.0307

Finally, we can find the frictional force acting on the smaller block:

F(friction) = uN 
F(friction) = (0.0307)(19.62 N)

F(friction) = 0.603 N

Since the question asks for the horizontal component of the force of the smaller block on the larger block, we can use trigonometry to find that:

Horizontal component = F(friction)cos(30°)

Horizontal component = (0.603 N)(cos 30°)

Horizontal component = 0.72 N to the right

Therefore, the answer is B.

Two Blocks Are Accelerated Across a Horizontal Frictionless Surface as Shown