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10P

Expert-verifiedFound in: Page 171

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**A coin slides over a frictionless plane and across an xy coordinate system from the origin to a point with xy coordinates ${\left(3.0m,4.0m\right)}$ while a constant force acts on it. The force has magnitude 2.0 N**** ****and is directed at a counterclockwise angle of ${{\mathbf{100}}}^{{\mathbf{\xb0}}}$ from the positive direction of the x axis. How much work is done by the force on the coin during the displacement?**

6.8 J of work is done by the force of 2.0 N on a coin during displacement.

The initial position of the coin$=\left(0,0\right)$.

The final position of the coin $=\left(3.0\mathrm{m},4.0\mathrm{m}\right)$.

Force is, $\mathrm{F}=20\mathrm{N}$.

The angle at which the force is directed is, $\mathrm{\theta}={100}^{\xb0}$.

**Using the formula for work in terms of force and displacement, the work done in the x and y directions can be found separately. After this, we can add them to get the total work done by the force on the coin.**

Formula:

$\mathrm{W}=\mathrm{F}\u2206\mathrm{x}\phantom{\rule{0ex}{0ex}}{\mathrm{W}}_{\mathrm{total}}={\mathrm{W}}_{\mathrm{x}}+{\mathrm{W}}_{\mathrm{y}}$

A free body diagram can be drawn as,

From the above diagram, we can calculate the work done in the x direction as,

${\mathrm{W}}_{\mathrm{x}}={\mathrm{F}}_{\mathrm{x}}\u2206\mathrm{x}$

Here, $\u2206\mathrm{x}$ is the displacement in the x direction.

Substitute the values in the above expression, and we get,

${\mathrm{W}}_{\mathrm{x}}=\left(2.0\mathrm{N}\times \mathrm{cos}\left({100}^{\xb0}\right)\right)\times \left(3.0\mathrm{m}-0\mathrm{m}\right)\phantom{\rule{0ex}{0ex}}{\mathrm{W}}_{\mathrm{x}}=-1.01\mathrm{N}.\mathrm{m}\phantom{\rule{0ex}{0ex}}{\mathrm{W}}_{\mathrm{x}}=-1.01\mathrm{J}$

From the above diagram, we can calculate the work done in the y direction as,

${\mathrm{W}}_{\mathrm{y}}={\mathrm{F}}_{\mathrm{y}}\u2206\mathrm{y}$

Here, $\u2206\mathrm{y}$ is the displacement in the y direction.

Substitute the values in the above expression, and we get,

${\mathrm{W}}_{\mathrm{y}}=\left(2\mathrm{N}\times \mathrm{sin}\left({100}^{\xb0}\right)\right)\times \left(4.0\mathrm{m}-0\mathrm{m}\right)\phantom{\rule{0ex}{0ex}}{\mathrm{W}}_{\mathrm{y}}=7.88\mathrm{N}.\mathrm{m}\phantom{\rule{0ex}{0ex}}{\mathrm{W}}_{\mathrm{y}}=7.88\mathrm{J}$

Total work done will be,

$\mathrm{W}={\mathrm{W}}_{\mathrm{x}}+{\mathrm{W}}_{\mathrm{y}}$

Substitute the values in the above expression, and we get,

$\mathrm{W}=-1.01\mathrm{J}+7.88\mathrm{J}\phantom{\rule{0ex}{0ex}}\mathrm{W}=6.8\mathrm{J}$

6.8 J of work is done by the force of 2.0 N on a coin during displacement.

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