In the figure, a 700-kg crate is on a rough surface inclined at 30°. a constant external force p = 5600 n is applied horizontally to the crate. as the force pushes the crate a distance of 3.00 m up the incline, the speed changes from 1.40 m/s to 2.50 m/s. how much work does gravity do on the crate during this process?

This problem might be confusing since there are a lot of given variable but we must take note that the problem only asks about the Work done by gravity, therefore:

Work = - Potential Energy

Where,

Potential Energy = mass * gravity * height

Note: It is negative since it is gravity doing the work on our system, not work being done on gravity.

Since the surface is inclined at 30 degrees, therefore:

height = 3 m * sin 30

height = 1.5 m

Calculating for Work:

Work = - 700 kg * 9.8 m / s^2 * 1.5 m

Work = - 10,290 J

Work = -10,300 J

Work do on the crate by gravity : W = -10290 J

Work is the transfer of energy caused by the force acting on a moving object

Work is the product of force with the displacement of objects.

Can be formulated

W = Work, J, Nm

F = Force, N

S = distance, m

Force and displacement of objects are vector. Multiplication of both is in accordance with the dot product concept of two vectors. Whereas W is a scalar quantity. Then if there is an angle formed by the force F and the displacement S is θ, then it can be formulated

There are 4 forces on the crate, so there are 4 work that occurs

the magnitude is the same as the force applied by the applied force (Fp) but the opposite direction

the direction is upward with F = mg

Because what is sought is only work by the force of gravity, then

we can find it from the formula of potential energy

W = ΔPE = - m. g. (h2-h1)

W = - mgh2 (h1 = 0)

Negative value are caused by the movement of objects in the opposite direction to the downward force of gravity

W = - 700. 9.8. 3 sin 30 (h2 has an angle of 30°)

W = -10290 J

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