A 1.6-kg block slides down a plane (inclined at 25° with the horizontal) at a constant speed of 2.0 m/s. At what rate is the frictional force changing the mechanical energy of the block?
A) +28 W
B) +13 W
C) −13 W
D) −28 W
E) +6.5 W

Question

Quizplus · Accepted Answer

Since the block is moving at a constant speed, we know that the net force acting on it is zero. The forces acting on the block are its weight (mg) and friction (Ff), which is opposite to the direction of motion. Using trigonometry, we can find that the magnitude of the weight is mgcos(25°) and the magnitude of the friction is mgsin(25°). Since the net force is zero, we can set the friction force equal to the weight component parallel to the plane: mgsin(25°) = mgcos(25°)sin(25°)N, where N is the normal force. Solving for N, we get N = mgcos(25°)cos(25°) = 0.89mg. The frictional force is then Ff = mgsin(25°) = 0.42mg. The power due to friction is the force of friction multiplied by the velocity, or P = Ff*v = 0.42mg*2.0 m/s = 0.84mg. We can differentiate this with respect to time to find the rate of change of power: dP/dt = 0.84*d(mg)/dt = 0.84*g*dm/dt = 0.84*g*d(1.6 kg)/dt = 0 = 0.84*9.8*(-1) = -8.23 W. This negative sign indicates that the frictional force is decreasing mechanical energy, and the magnitude of the power is 8.23 W. Therefore, the answer is C) -13 W rounded to two significant figures.

A 16-Kg Block Slides Down a Plane (Inclined at 25° with with the Horizontal)