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Q105P

Expert-verifiedFound in: Page 211

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 ${\mathbf{40}}{\mathbf{\xb0}}$**** 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?**

- The work done on the trunk by the machine’s force is 744 J
- The increase in thermal energy of the trunk and the ramp is 240.2 J

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, $\theta =40\xb0$** **

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

**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)

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:

${\mathrm{F}}_{\mathrm{m}}=\mathrm{mgsin\theta}+{\mathrm{\mu}}_{\mathrm{k}}\mathrm{mgcos\theta}\phantom{\rule{0ex}{0ex}}=40\mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^{2}\times \left(\mathrm{sin}40\xb0\right)+0.40\times \mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^{2}\times \left(\mathrm{cos}40\xb0\right)\phantom{\rule{0ex}{0ex}}=251.9\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^{2}+120.1\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^{2}\phantom{\rule{0ex}{0ex}}=372\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^{2}\left(\frac{1\mathrm{N}}{1\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^{2}}\right)$

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

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

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

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:

${\mathrm{WD}}_{\mathrm{th}}={\mathrm{\mu}}_{\mathrm{k}}\mathrm{mgcos\theta}\times \mathrm{d}\phantom{\rule{0ex}{0ex}}=0.40\times 40\mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^{2}\times \mathrm{cos}40\xb0\times 2.0\mathrm{m}\phantom{\rule{0ex}{0ex}}=240.2\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^{2}\left(\frac{1\mathrm{J}}{1\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^{2}}\right)\phantom{\rule{0ex}{0ex}}=240.2\mathrm{J}$

Hence, the value of the increase in thermal energy is 240.2 J .

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