As the crate slides 3.6m, how much work is done on the crate by: a) the worker b) the force of gravity c) the normal force due to the incline plane?

Question

1. To push a 25kg up a 27* inclined plane, a worker exerts a force of 120N parallel to the inclined plane. As the crate slides 3.6m, how much work is done on the crate by:

a) the worker

b) the force of gravity

c) the normal force due to the incline plane?

Answer 1

Hello Llama!

I think it is important to first define the problem with a picture and your knowns.

KNOWNS: Θ = 27^o, m_crate = 25kg, F_push=110N, and d=3.6m

Using basic trigonometry, we can see that:

Δh = 3.6sin(27^o)

Δx = 3.6cos(27^o)

We also know that: W = F * d, and work done in the same direction as motion is positive while opposite direction is negative.

If we separate this into a problem where the coordinate system is defined with the x-direction pointing left and the y-direction pointing up, then we can look at the x-dir and y-dir separately:

y-dir:

W_gravity = F_gravity*d_y = mgΔh (negative)

W_normal = F_normal*d_y = F_Ncos(27^o)Δh (positive)

W_worker = F_push*d_y = F_pushsin(27^o)Δh (positive)

x-dir:

W_gravity = F_gravity*d_y = mgΔh = 0 (there is no x-component to the gravitational force)

W_normal = F_normal*d_x = F_Nsin(27^o)Δx (negative)

W_worker = F_push*d_x = F_pushcos(27^o)Δx (positive)

Now just add Wx and Wy to get the final result:

W_gravity = 0 + mgΔh

W_normal = F_Nsin(27^o)Δx + F_Ncos(27^o)Δh

W_worker = F_pushcos(27^o)Δx + F_pushsin(27^o)Δh

Don't forget to use the correct signs in your calculations. Also note, conceptually, if you have done it the right way your result should equal this as well:

W_worker = F_pushcos(27^o)Δx + F_pushsin(27^o)Δh = F_push * Δd = 120N * 3.6m = 432J (or Nm)