# 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 ?

Nov 10, 2017

$2 \cdot {10}^{2} \text{ J}$

#### Explanation:

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

$W = {\int}_{{s}_{0}}^{{s}_{f}} F \cdot \mathrm{ds}$

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):

$W = F \left({s}_{f} - {s}_{0}\right)$

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 $m g \sin \left(\theta\right)$, 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:

$\Sigma {F}_{x} = m {a}_{x}$
x signifies the direction up the ramp

$\implies {F}_{p} - m g \sin \left(\theta\right) = 0$

Now, we just solve for ${F}_{p}$:

${F}_{p} = m g \sin \left(\theta\right) = \left(3\right) \left(9.8\right) \left(\sin \left(60\right)\right)$

$\implies {F}_{p} = 25.5 N$

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

$W = F \left({s}_{f} - {s}_{0}\right)$

We can set ${s}_{0} = 0$, and ${s}_{f} = 8$:

$W = 25.5 \left(8 - 0\right)$

$\implies W = 203.6 \to 2 \cdot {10}^{2} \text{ J}$.
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 :)