Question

How much work would it take to push a `3 kg` weight up a `8 m` plane that is at an incline of `pi / 3`?

Answer 1

`2 * 10^2 " J"`

Explanation:

We know that the technical, calculus-based definition of work is defined as:

`W = int_(s_0)^(s_f)F * ds`

where `s` is position.

Assuming a constant force, this expression simplifies to the following (which you are taught if you're not doing a calculus based physics course):

`W = F(s_f - s_0)`

We know how far the block travels, but we do not know how much force is acting on the block. Hence, we'll need to set up a free body diagram and use Newton's 2nd Law to figure that out:

Before we start talking about work, let's draw a free body diagram here to analyze the forces acting on the crate as we push it up:

I've assumed that the ramp is frictionless, since taking friction into account would need a bit more information that you don't have.

Hence, the only forces we care about here are those acting along the axis of the ramp. In this case, those are `F_p`, our pushing force (which we're solving for), and `mgsin(theta)`, a component of gravitational force.

Recall that in order to get the ramp moving, the minimum net force you need to have acting on the object is 0 (meaning that the object would have zero acceleration). Therefore, we can set up the following:

`SigmaF_x = ma_x`
x signifies the direction up the ramp

`=> F_p - mgsin(theta) = 0`

Now, we just solve for `F_p`:

`F_p = mgsin(theta) = (3)(9.8)(sin(60))`

`=> F_p = 25.5 N`

That is the minimum force we need to move the block. Now, we can go ahead and set up our work equation:

`W = F(s_f - s_0)`

We can set `s_0 = 0`, and `s_f = 8`:

`W = 25.5(8 - 0)`

`=> W = 203.6 -> 2 * 10^2 " J"`.
Rounded to one significant figure

And there you are. Note that this is the minimum amount of work you'd have to do. You could apply a larger force on the block, and move it up faster, but you'd be doing more work in those cases.

Hope that helped :)