A 3.0-kg block moves up a 40° incline with constant speed under the action of a 26-N force acting up and parallel to the incline. What magnitude force must act up and parallel to the incline for the block to move down the incline at constant velocity?
A) 14 N
B) 12 N
C) 16 N
D) 18 N
E) 25 N

Question

Quizplus · Accepted Answer

When the block moves up the incline with constant speed, the force of gravity acting down the incline is balanced by the component of the force parallel to the incline. Since there is no acceleration, the net force on the block must be zero.
Therefore, the force of gravity acting down the incline is given by:

$F_g=mg\sin(	heta)=3.0	ext{ kg}	imes9.8	ext{ m/s}^2	imes\sin(40°)=18.9	ext{ N}$

The 26-N force acting up and parallel to the incline must therefore be balanced by a force acting down the incline with magnitude:

$F_{down}=26	ext{ N}-18.9	ext{ N}=7.1	ext{ N}$

When the block moves down the incline with constant velocity, the force of gravity acting down the incline is balanced by the component of the force parallel to the incline minus the force acting up and parallel to the incline. Since there is no acceleration, the net force on the block must be zero.

Therefore, the force of gravity acting down the incline is still given by:

$F_g=18.9	ext{ N}$

The force acting down the incline parallel to the incline must therefore be:

$F_{down}=F_g-F_{up}=18.9	ext{ N}-7.1	ext{ N}=11.8	ext{ N}$

and this force is acting in the opposite direction compared to the 26-N force in the previous case. Using trigonometry, we find that the force parallel to the incline required to balance the forces and maintain constant velocity is:

$F_{up}=F_{down}\sin(	heta)=11.8	ext{ N}\sin(40°)=8.0	ext{ N}$

The force acting up and parallel to the incline must therefore have a magnitude of:

$F_{up}=26	ext{ N}-8.0	ext{ N}=18	ext{ N}$

so the answer is B.

A 30-Kg Block Moves Up a 40° Incline with Constant Speed