# How much work does it take to push an object with a mass of 2 kg up a 2 m ramp, if the ramp has an incline of (11pi)/12  and a kinetic friction coefficient of 3 ?

Nov 21, 2017

My calculator is in the degree mode, so I will convert $\frac{11 \pi}{12} {\text{to}}^{\circ}$.

$\frac{11 \cancel{\pi}}{12} \cdot {180}^{\circ} / \cancel{\pi} = {165}^{\circ}$

Note: $\frac{\pi}{2} = {90}^{\circ}$ so this ramp is inclined so far that it is almost upside down. Another ${15}^{\circ}$ and it would be level.

I prefer to use conservation of energy. The top of the ramp has gravitational potential energy of
$m \cdot g \cdot h = 2 k g \cdot 9.8 \frac{m}{s} ^ 2 \cdot 2 m \cdot \sin 165 = 10.1 k g \cdot {m}^{2} / {s}^{2} = 10.1 J$

In addition to that, work was done against the force of friction -- call it ${F}_{f}$. That work is
$\text{work} = {F}_{f} \cdot s = {\mu}_{k} \cdot N \cdot s = 3 \cdot m \cdot g \cdot \cos 165 \cdot 2 m$
$\text{work" = 114" } k g \cdot {m}^{2} / {s}^{2} = 114 J$

So the sum is $10.1 J + 114 J \cong 124 J$

Now I will calculate work using the force *distance method. (Because cos165 has a negative value, I will use cos15 instead of cos165 in the following steps.)
The force of friction from the work above is
${F}_{f} = 3 \cdot m \cdot g \cdot \cos 15 = 3 \cdot 2 k g \cdot 9.8 \frac{m}{s} ^ 2 \cdot \cos 15$
${F}_{f} = 56.8 k g \cdot \frac{m}{s} ^ 2 = 56.8 N$

Exerting that force over the 2 m length calculates to $2 m \cdot 56.8 N = 114 J$
There is also some work against gravity. The downslope component of the weight of that mass, ${W}_{\mathrm{ds}}$ is
${W}_{\mathrm{ds}} = m \cdot g \cdot \sin 165 = 2 k g \cdot 9.8 \frac{m}{s} ^ 2 \cdot \sin 165 = 5.07 N$

So the work of pushing against that force is $5.07 N \cdot 2 m = 10.14 J$

So the sum is $10.1 J + 114 J \cong 124 J$

I hope this helps,
Steve