Question

A tension force of 165 N inclined at 25.0° above the horizontal is used to pull a 37.0 kg packing crate a distance of 3.60 m on a rough surface. If the crate moves at a constant speed, find (a) the work done by the tension force and (b) the coefficient of kinetic friction between the crate and surface.

a) (a) 542.52 J (b) 0.3
b) (a) 325.80 J (b) 0.2
c) (a) 451.60 J (b) 0.4
d) (a) 678.90 J (b) 0.5

Answer 1

The work done by the tension force is zero, and the coefficient of kinetic friction between the crate and the surface is approximately 0.3. option a is the correct answer.

Step-by-step explanation:

To solve this problem, we can use the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy. Since the crate is moving at a constant speed, its kinetic energy remains the same. Therefore, the work done by the tension force is zero (a).

To find the coefficient of kinetic friction (b), we can use the equation:

μk = (work done by friction) / (normal force * distance)

First, let's calculate the normal force:

Normal force = mass * gravity

Normal force = 37.0 kg * 9.8 m/s^2

Normal force = 362.6 N

Next, we can use the equation:

(work done by friction) = (force of friction) * (distance)

Since the crate is moving at a constant speed, the work done by friction is equal to the work done by the tension force:

(work done by friction) = 165 N * 3.60 m

(work done by friction) = 594 J

Finally, we can substitute the values into the equation:

μk = 594 J / (362.6 N * 3.60 m)

μk ≈ 0.3 (rounded to one decimal place)

Therefore, the correct option is (a) 542.52 J (b) 0.3.