A 14.7 kg crate is pulled up a rough incline

with an initial speed of 1.5 m/s. The pulling
force is 124.0 N parallel to the incline, which
makes an angle of 11.0

with the horizontal.
The coefficient of kinetic friction is 0.31 and
the crate is pulled a distance of 7.2 m.
The acceleration of gravity is 9.81 m/s
2
.
a) Find the work done by Earth’s gravity
on the crate.
Answer in units of J

Fc = m*g = 14.7kg * 9.81N/kg = 144.1 N.=

Force of crate.

d = 7.2*sin11 = 1.37 m., Vertical.

Work = Fc * d = 144.1 * 1.37=198 Joules.

To find the work done by Earth's gravity on the crate, we need to use the formula:

Work = force * distance * cos(theta)

Where:
- force is the weight of the crate, which is given by the mass (14.7 kg) multiplied by the acceleration due to gravity (9.81 m/s^2).
- distance is the distance the crate is pulled (7.2 m).
- theta is the angle between the force of gravity and the direction of motion, which is 0 degrees since gravity acts vertically downward.

Let's calculate it step by step:

First, calculate the weight of the crate:
Weight = mass * acceleration due to gravity = 14.7 kg * 9.81 m/s^2

Next, calculate the work done by gravity:
Work = Weight * distance * cos(theta)

Since cos(0) = 1, we can simplify the formula to:
Work = Weight * distance

Finally, substitute the values into the formula and calculate the answer.