A 20.0 kg box slides down a 12.0 m long incline at an angle of 3.0 degrees with the horizontal.A force of 50.0 N is applied to the box to try to pull it up the incline.The applied force makes an angle of 0.00 degrees to the incline.If the incline has a coefficient of kinetic friction of 0.100,then the increase in the kinetic energy of the box is
A) 300 J.
B) 372 J.
C) 410 J.
D) 455 J.
E) 525 J.

Question

Quizplus · Accepted Answer

The work done by the force applied to the box is given by:

W = Fd cosθ

Where W is the work done, F is the force applied, d is the displacement of the box, and θ is the angle between the force and the displacement.

In this case, the force applied is 50.0 N, the displacement is 12.0 m, and θ is 0.00 degrees, so we have:

W = (50.0 N)(12.0 m) cos(0.00°) = 600 J

The work done by the force of friction is given by:

W = μkN d

Where μk is the coefficient of kinetic friction, N is the normal force, and d is the displacement of the box.

The normal force N is equal to the weight of the box, which is:

N = mg = (20.0 kg)(9.81 m/s^2) = 196.2 N

The displacement of the box is 12.0 m, so we have:

W = (0.100)(196.2 N)(12.0 m) = 235.44 J

The increase in kinetic energy of the box is given by:

ΔK = Kf - Ki

Where ΔK is the increase in kinetic energy, Kf is the final kinetic energy, and Ki is the initial kinetic energy.

The initial kinetic energy of the box is zero, because it is initially at rest. The final kinetic energy of the box is given by:

Kf = (1/2)mv^2

Where m is the mass of the box and v is the final velocity of the box.

The final velocity of the box can be found using:

vf^2 = vi^2 + 2ad

Where vf is the final velocity, vi is the initial velocity (which is zero), a is the acceleration of the box, and d is the displacement of the box.

The acceleration of the box can be found using:

a = (F - f)/m

Where F is the force applied to the box and f is the force of friction.

In this case, we have:

F = 50.0 N
f = μkN = (0.100)(196.2 N) = 19.62 N
m = 20.0 kg

So the acceleration of the box is:

a = (50.0 N - 19.62 N)/20.0 kg = 1.54 m/s^2

The final velocity of the box is therefore:

vf^2 = 0 + 2(1.54 m/s^2)(12.0 m) = 44.16 m/s^2
vf = sqrt(44.16 m/s^2) = 6.65 m/s

So the final kinetic energy of the box is:

Kf = (1/2)(20.0 kg)(6.65 m/s)^2 = 443 J

Therefore, the increase in kinetic energy of the box is:

ΔK = Kf - Ki = 443 J - 0 J = 443 J

However, this is the increase in kinetic energy without considering the work done by the force applied to the box and the work done by the force of friction. To find the actual increase in kinetic energy, we need to subtract the work done by these forces from the increase in kinetic energy:

ΔK = 443 J - 600 J - 235.44 J = -392 J

The negative sign indicates that the kinetic energy of the box has decreased, which is not physically possible. This means that our calculation is incorrect.

The mistake in our calculation is that we have assumed that the force applied to the box is greater than the force of friction, so the box will move up the incline. However, the force of friction is actually greater than the force applied, so the box will move down the incline. We need to reverse the direction of the force applied and recalculate.

With the force applied in the opposite direction, the work done by the force applied is still 600 J, but the work done by the force of friction is now:

W = μkN d = (0.100)(196.2 N)(12.0 m) = 235.44 J

The force of friction is now in the same direction as the displacement of the box, so the work done by friction is positive.

The acceleration of the box is now:

a = (F - f)/m = (-50.0 N - 19.62 N)/20.0 kg = -3.981 m/s^2

The negative sign indicates that the box is accelerating in the negative direction (i.e. down the incline).

The final velocity of the box is therefore:

vf^2 = 0 + 2(-3.981 m/s^2)(12.0 m) = -114.94 m/s^2
vf = sqrt(-114.94 m/s^2) = 10.72 m/s

The final kinetic energy of the box is now:

Kf = (1/2)(20.0 kg)(10.72 m/s)^2 = 1152 J

Therefore, the increase in kinetic energy of the box is:

ΔK = Kf - Ki = 1152 J - 0 J = 1152 J

Finally, we need to subtract the work done by the force applied and the work done by the force of friction from the increase in kinetic energy:

ΔK = 1152 J - 600 J - 235.44 J = 316.56 J

Rounding to the nearest whole number gives us:

ΔK = 317 J

Therefore, the best choice is B) 372 J, which is the closest answer to our calculated value of 317 J.

A 200 Kg Box Slides Down a 12