Physics, 04.11.2019 20:31 spaigenicole

Acart for hauling ore out of a gold mine has a mass of 413 kg , including its load. the cart runs along a straight stretch of track that climbs a shallow 4.09° incline. a donkey, which is trudging along and to the side of the track, has the unenviable job of pulling the cart up the slope with a 366 n force for a distance of 163 m by means of a rope that is parallel to the slope but makes an angle of 14.7° with the track. the coefficient of friction for the cart\'s wheels on the track is 0.0105. take g = 9.81 m/s2. find the work that the donkey performs on the cart and gravity performs on the cart during this process. as well as calculate the work done on the cart during the process by friction.
(a)-donkey work on cart:
(b)-gravity work on cart:
(c)-friction work on cart:

Answers: 3

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Answer from: helpplz5725

We have that from the Question, it can be said that

(a)-Donkey work on cart: Wg = Wd=57705.259J

(b)-Gravity work on cart: Wg = 47101.9J

(c)-Friction work on cart: Wf =7284.969 J

From the Question we are told

A cart for hauling ore out of a gold mine has a mass of 413 kg , including its load. The cart runs along a straight stretch of track that climbs a shallow 4.09° incline. a donkey, which is trudging along and to the side of the track, has the unenviable job of pulling the cart up the slope with a 366 n force for a distance of 163 m by means of a rope that is parallel to the slope but makes an angle of 14.7° with the track. The coefficient of friction for the cart\'s wheels on the track is 0.0105. Take g = 9.81 m/s2. Find the work that the donkey performs on the cart and gravity performs on the cart during this process. As well as calculate the work done on the cart during the process by friction.

(a)-Donkey work on cart:

(b)-Gravity work on cart:

(c)-Friction work on cart:

Generally the equation for Donkey work on cart is mathematically given as

Wd = F d cos(theta)

Wd=366*163cos14.7

Wd=57705.259J

Generally the equation for Gravity work on cart: is mathematically given as

Wg = m g h

Wg = 413 * 9.81 *163 sin(4.09)

Wg = 47101.9J

Generally the equation for Friction work on cart: is mathematically given as

Wf = \mu m g cos(theta)d

The

Wf = 0.0105 * 435 * 9.81 * cos(4.09) * 163

Wf =7284.969 J

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Answer from: ehehvevsbg

(a) 57700 J

(b) -47120 J

Explanation:

Component of force applied by the donkey in the direction of movement is given by

${F_{\rm{d}}} = F\left( {\cos ({\theta _1})} \right)$

Substitute $366\,{\rm{N}$ for F and $14.7^\circ$ for $\theta$ .

$\begin{array}{c}\\{F_{\rm{d}}} = \left( {366\,{\rm{N}}} \right)\left( {\cos (14.7^\circ )} \right)\\\\ = 354.{\rm{02}}\,{\rm{N}}\\\\ \approx {\rm{354}}\,{\rm{N}}\\\end{array}$

Work done by the donkey on the cart,

${W_{\rm{d}}} = {F_{\rm{d}}}$

Substitute $354\,{\rm{N}}$ for ${F_{\rm{d}}}$ and $163\,{\rm{m}}$ for d .

$\begin{array}{c}\\{W_{\rm{d}}} = \left( {354\,{\rm{N}}} \right)\left( {163\,{\rm{m}}} \right)\\\\ = 57702\,{\rm{J}}\\\\ \approx {\rm{57700}}\,{\rm{J}}\\\end{array}$

Change in the height of the cart from the horizontal is,

$h = d\sin \theta$

Substitute $163\,{\rm{m}}$ for d and $4.09^\circ$ for $\theta$ .

$\begin{array}{c}\\h = \left( {163\,{\rm{m}}} \right)\left( {\sin \left( {4.09^\circ } \right)} \right)\\\\ = 11.62571\,{\rm{m}}\\\\ \approx {\rm{11}}{\rm{.63}}\,{\rm{m}}\\\end{array}$

Work done by gravity is given as,

${W_{\rm{g}}} = - mgh$

Substitute $413\,{\rm{kg}}$ for m, $9.81\,{\rm{m/}}{{\rm{s}}^2}$ for g, and $11.63\,{\rm{m}}$ for h.

$\begin{array}{c}\\{W_{\rm{g}}} = - \left( {413\,{\rm{kg}}} \right)\left( {9.81\,{\rm{m/}}{{\rm{s}}^2}} \right)\left( {11.63\,{\rm{m}}} \right)\\\\ = - 47119.29\,{\rm{J}}\\\\ \approx - 47120\,{\rm{J}}\\\end{array}$

Force of friction is given by

${F_{\rm{f}}} = - {\mu _{\rm{f}}}\left( {mg\cos (\theta )} \right)$

Substitute 0.0105 for ${\mu _{\rm{f}}}$ , $413\,{\rm{kg}}$ for m, $9.81\,{\rm{m/}}{{\rm{s}}^2}$ for g, and $4.09^{o}$ for $\theta$ .

$\begin{array}{c}\\{F_{\rm{f}}} = - \left[ {\left( {0.0105} \right)\left( {\left( {413\,{\rm{kg}}} \right)\left( {9.81\,{\rm{m/}}{{\rm{s}}^2}} \right)\cos (4.09^\circ )} \right)} \right]\\\\ = - 42.42187\,{\rm{N}}\\\\ \approx {\rm{-42}}{\rm{.4}}\,{\rm{N}}\\\end{array}$

The expression for the work done by the frictional force is,

${W_{\rm{f}}} = {F_{\rm{f}}}$

Substitute $42.4\,{\rm{N}}$ for ${F_{\rm{f}}}$ and $163\,{\rm{m}$ }for d.

$\begin{array}{c}\\{W_{\rm{f}}} = - \left( {42.4\,{\rm{N}}} \right)\left( {163\,{\rm{m}}} \right)\\\\ = - 6911.2\,{\rm{J}}\\\\ \approx - {\rm{6911}}\,{\rm{J}}\\\end{array }$

Answer from: Quest

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Answer from: Quest

No because scientist are still changing there mines today so they are not correct to this day

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Acart for hauling ore out of a gold mine has a mass of 413 kg , including its load. the cart runs al...