In the figure,a 900-kg crate is on a rough surface inclined at 30°.A constant external force is applied horizontally to the crate.While this force pushes the crate a distance of 3.0 m up the incline,its velocity changes from 1.2 m/s to 2.3 m/s.How much work does friction do during this process?
A) -3700 J
B) -7200 J
C) +3700 J
D) +7200 J
E) zero

Question

In the figure,a 900-kg crate is on a rough surface inclined at 30°.A constant external force  is applied horizontally to the crate.While this force pushes the crate a distance of 3.0 m up the incline,its velocity changes from 1.2 m/s to 2.3 m/s.How much work does friction do during this process? 
A) -3700 J 
B) -7200 J 
C) +3700 J 
D) +7200 J 
E) zero

Quizplus · Accepted Answer

We can use the work-energy theorem to solve this problem. The work-energy theorem states that the net work done on an object is equal to its change in kinetic energy.

First, we need to find the change in the crate's kinetic energy:

ΔKE = ½ mv₂² - ½ mv₁²
ΔKE = ½ (900 kg)(2.3 m/s)² - ½ (900 kg)(1.2 m/s)²
ΔKE = 2079 J

Next, we need to find the net work done on the crate. This is equal to the work done by the external force minus the work done by friction:

Wnet = Wext - Wfr

We can find Wext by multiplying the force by the distance it acts over:

Wext = Fd cosθ
Wext = (110 N)(3.0 m) cos(30°)
Wext = 891 J

To find Wfr, we need to use the fact that the force of friction is equal to the coefficient of friction multiplied by the normal force (which is equal to the weight of the crate):

f fr = μ fr N = μ fr mg

The normal force is perpendicular to the incline and is equal to:

N = mg cosθ
N = (900 kg)(9.81 m/s²) cos(30°)
N = 7350 N

Therefore, the force of friction is:

f fr = μ fr N
f fr = (0.2)(7350 N)
f fr = 1470 N

The work done by friction is equal to the force of friction multiplied by the distance it acts over:

Wfr = f fr d cos180°
Wfr = (1470 N)(3.0 m) cos(180°)
Wfr = - 4410 J

Note that we have included the negative sign because friction acts in the opposite direction of motion.

Finally, we can substitute the values we found into the work-energy equation:

Wnet = ΔKE
Wext - Wfr = ΔKE
891 J - (-4410 J) = 2079 J
Wnet = - 3700 J

Therefore, the work done by friction is -3700 J, which means the answer is A.

In the Figure,a 900-Kg Crate Is on a Rough Surface