Pete slides a crate up a ramp at an angle of 20 degrees by exerting a 244 N force parallel to the ramp. The crate moves at a constant speed. The coefficient of friction is 0.19. How much work did Pete do when the crate was raised a vertical distance of 2.39 m? Answer in units of J

Drwls is certainly not correct.

To find the work done by Pete in raising the crate, we need to calculate the total work done against gravity and friction.

First, let's determine the component of the force parallel to the ramp, which is responsible for raising the crate. This can be found by multiplying the force Pete exerted (244 N) by the cosine of the angle of the ramp (20 degrees):

Force parallel to the ramp = 244 N * cos(20°)

Next, we can calculate the work done against gravity. The work done against gravity is equal to the force magnitude multiplied by the vertical distance the crate was raised. In this case, the vertical distance is given as 2.39 m. Since the crate moves at a constant speed, there is no change in potential energy, and thus the work done against gravity is equal to the work done by gravity:

Work against gravity = -m * g * h

where m is the mass of the crate, g is the acceleration due to gravity (9.8 m/s^2), and h is the vertical distance (2.39 m).

Finally, we need to calculate the work done against friction. The force of friction can be found by multiplying the coefficient of friction (0.19) by the normal force, which is equal to the weight of the crate:

Force of friction = coefficient of friction * Normal force

Since the crate is on an incline, the normal force can be found by multiplying the mass of the crate by the acceleration due to gravity (9.8 m/s^2):

Normal force = m * g

Now we can find the work done against friction using the formula:

Work against friction = force of friction * distance

where the distance is equal to the vertical distance (2.39 m).

Once we have all three values, we can calculate the total work done by adding the work done against gravity and the work done against friction:

Total work done by Pete = Work against gravity + Work against friction

Plug in the values into the formulas and calculate to find the answer in units of joules (J).

Pete slides a crate up a ramp at an angle of

35◦ by exerting a 179 N force parallel to the
ramp. The crate moves at a constant speed.
The coefficient of friction is 0.21.
How much work did Pete do when the crate
was raised a vertical distance of 1.83 m?
Answer in units of J.

Force * distance = 244*(2.39/sin20)

= 1705 J

You do not need to know the coefficient of friction.