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Expert-verified Found in: Page 211 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # A machine pulls a 40 kg trunk 2.0 m up a ${\mathbf{40}}{\mathbf{°}}$ ramp at a constant velocity, with the machine’s force on the trunk directed parallel to the ramp. The coefficient of kinetic friction between the trunk and the ramp is 0.40 . What are (a) the work done on the trunk by the machine’s force and (b) the increase in thermal energy of the trunk and the ramp?

1. The work done on the trunk by the machine’s force is 744 J
2. The increase in thermal energy of the trunk and the ramp is 240.2 J
See the step by step solution

## Step 1: The given data

The mass of the trunk is, m = 40 kg

The distance by which the trunk is pulled up, x = 2.0 m

The angle of inclination is, $\theta =40°$

The coefficient of kinetic friction is, ${\mu }_{k}=0.40$

## Step 2: Understanding the concept of energy and force

Using the equation for the net force, we can find the force due to the machine. From this force and displacement, we can find the work done by the machine. Using the frictional force, we can find the thermal energy generated.

Formulae:

Force due to Newton’s second law, F = ma (1)

The work done by the body, W = Fdcos$\mathrm{\theta }$ (2)

## Step 3: a) Calculation of the work done on the trunk

We can resolve the gravitational force into its component along the incline and normal to the incline, we have and, respectively.

Using these components in equation (2), we get that

${\mathrm{F}}_{\mathrm{m}}-\mathrm{mgsin\theta }-{\mathrm{\mu }}_{\mathrm{k}}\mathrm{mgcos\theta }=\mathrm{ma}$

The net acceleration is zero, so we can write the force as:

${\mathrm{F}}_{\mathrm{m}}=\mathrm{mgsin\theta }+{\mathrm{\mu }}_{\mathrm{k}}\mathrm{mgcos\theta }\phantom{\rule{0ex}{0ex}}=40\mathrm{kg}×9.8\mathrm{m}/{\mathrm{s}}^{2}×\left(\mathrm{sin}40°\right)+0.40×\mathrm{kg}×9.8\mathrm{m}/{\mathrm{s}}^{2}×\left(\mathrm{cos}40°\right)\phantom{\rule{0ex}{0ex}}=251.9\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^{2}+120.1\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^{2}\phantom{\rule{0ex}{0ex}}=372\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^{2}\left(\frac{1\mathrm{N}}{1\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^{2}}\right)$

Thus, work done by the machine can be given as:

$\mathrm{W}={\mathrm{F}}_{\mathrm{m}}×\mathrm{d}\phantom{\rule{0ex}{0ex}}=372\mathrm{N}×2\mathrm{m}\phantom{\rule{0ex}{0ex}}=744\mathrm{N}.\mathrm{m}\left(\frac{1\mathrm{J}}{1\mathrm{N}.\mathrm{m}}\right)\phantom{\rule{0ex}{0ex}}=744\mathrm{J}$

Hence, the value of the work done is 744J.

## Step 4: b) Calculation of the increased thermal energy

The force due to friction is given as:

${\mathrm{F}}_{\mathrm{f}}={\mathrm{\mu }}_{\mathrm{k}}\mathrm{mgcos\theta }$

So the thermal energy generated is equal to the work done by the frictional force, which is given using equation (2) as:

${\mathrm{WD}}_{\mathrm{th}}={\mathrm{\mu }}_{\mathrm{k}}\mathrm{mgcos\theta }×\mathrm{d}\phantom{\rule{0ex}{0ex}}=0.40×40\mathrm{kg}×9.8\mathrm{m}/{\mathrm{s}}^{2}×\mathrm{cos}40°×2.0\mathrm{m}\phantom{\rule{0ex}{0ex}}=240.2\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^{2}\left(\frac{1\mathrm{J}}{1\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^{2}}\right)\phantom{\rule{0ex}{0ex}}=240.2\mathrm{J}$

Hence, the value of the increase in thermal energy is 240.2 J . ### Want to see more solutions like these? 