A crate of mass 40 kg is pushed up a ramp by a person as shown in the figure below. The person pushes the crate a distance of 14 m as measured along the ramp. Assume the crate moves at constant velocity. Assume the coefficient of kinetic friction between the crate and the ramp is

μK = 0.35.
Angle between the horizontal and the ramp &=30°
Find the work done by gravity on the crate.

I tried to solve this for minutes but i can't find the right answer

To find the work done by gravity on the crate, we need to calculate the gravitational force acting on the crate and then multiply it by the distance the crate is moved.

First, let's calculate the gravitational force acting on the crate. The force of gravity can be calculated using the formula:

F_gravity = mass * gravity

Given that the mass of the crate is 40 kg, and the acceleration due to gravity is approximately 9.8 m/s^2, we can calculate the force of gravity:

F_gravity = 40 kg * 9.8 m/s^2 = 392 N

Next, we need to calculate the vertical component of the gravitational force. Since the crate is being pushed up the ramp, only a portion of the gravitational force is acting in the direction of the motion. The vertical component is given by:

F_vertical = F_gravity * sin(theta)

where theta is the angle between the horizontal and the ramp (which is 30° in this case).

F_vertical = 392 N * sin(30°) = 392 N * 0.5 = 196 N

Now, we can calculate the work done by gravity. The work done by a force is given by the formula:

Work = force * distance

Since the crate is being moved a distance of 14 m along the ramp, we can calculate the work done by gravity:

Work = F_vertical * distance = 196 N * 14 m = 2744 J

So, the work done by gravity on the crate is 2744 Joules.