A 3.0-kg block is dragged over a rough horizontal surface by a constant force of 16 N acting at an angle of 37° above the horizontal as shown.The speed of the block increases from 4.0 m/s to 6.0 m/s in a displacement of 5.0 m.What work was done by the friction force during this displacement?
A)−34 J
B)−64 J
C)−30 J
D)−94 J
E)+64 J

Question

A 3.0-kg block is dragged over a rough horizontal surface by a constant force of 16 N acting at an angle of 37° above the horizontal as shown.The speed of the block increases from 4.0 m/s to 6.0 m/s in a displacement of 5.0 m.What work was done by the friction force during this displacement? 
A)−34 J 
B)−64 J 
C)−30 J 
D)−94 J 
E)+64 J

Quizplus · Accepted Answer

First, we must use the work-energy principle, where the net work done on an object is equal to the object’s change in kinetic energy.

W_net = ΔK = (1/2)mv_f^2 - (1/2)mv_i^2

W_net = (1/2)(3.0 kg)(6.0 m/s)^2 - (1/2)(3.0 kg)(4.0 m/s)^2

W_net = 27 J

Next, we must find the work done by the 16 N force.

W_force = Fdcosθ
where θ is the angle between the displacement and the force.

W_force = (16 N)(5.0 m)cos(37°)

W_force = 61 J

Finally, we can find the work done by the friction force using the equation

W_friction = -W_net + W_force

W_friction = -(27 J) + (61 J)

W_friction = -34 J

Note that the negative sign indicates that the friction force is acting in the opposite direction of the displacement. Therefore, the best choice is (A), -34 J.

A 30-Kg Block Is Dragged Over a Rough Horizontal Surface by by a Constant