A 1.2-kg mass is projected up a rough circular track (radius = 0.80 m) as shown. The speed of the mass at point A is 8.4 m/s, and at point B, it is 5.6 m/s. What is the change in mechanical energy between A and B caused by the force of friction?

Question

Quizplus · Accepted Answer

The change in mechanical energy $\Delta E$ is given by

$$\Delta E = E_B - E_A = \frac{1}{2}mv_B^2 - \frac{1}{2}mv_A^2 - W_{	ext{fric}}$$

where $W_{	ext{fric}}$ is the work done by friction.

We can find the work done by friction as

$$W_{	ext{fric}} = \int_A^B \vec{f}_{	ext{fric}}\cdot d\vec{s}$$

where $\vec{f}_{	ext{fric}}$ is the frictional force, which acts in the opposite direction to the motion. Since the motion is upwards along the track, the frictional force points downwards, and its magnitude is given by

$$f_{	ext{fric}} = \mu mg\cos	heta$$

where $	heta$ is the angle between the track and the vertical. At point A, $	heta = 0$, so $f_{	ext{fric},A} = \mu mg$.

At point B, $	heta$ can be found using trigonometry:

$$	an	heta = \frac{h}{r} = \frac{r - (r\cos\phi)}{r} = 1 - \cos\phi$$

where $\phi$ is the angle between the track and the horizontal. Since the mass is moving upwards, we have $\phi > 90^\circ$, so $\cos\phi < 0$, and

$$\cos	heta = \frac{r - h}{r\sin\phi} = \frac{r - r\cos\phi}{r\sqrt{1-\cos^2\phi}} = \frac{1-\cos\phi}{\sqrt{1-\cos^2\phi}}$$

Substituting this into the expression for the frictional force, we get

$$f_{	ext{fric},B} = \mu mg\frac{1-\cos\phi}{\sqrt{1-\cos^2\phi}}$$

To evaluate the integral for the work done by friction, we need to express $d\vec{s}$ in terms of $d	heta$, since the motion is along the track. We have

$$d\vec{s} = -r\sin	heta d	heta\,\hat{r}$$

where $\hat{r}$ is a unit vector pointing from the center of the track towards the mass. The dot product $\vec{f}_{	ext{fric}}\cdot d\vec{s}$ is then

$$\vec{f}_{	ext{fric}}\cdot d\vec{s} = -f_{	ext{fric}}r\sin	heta d	heta$$

Substituting in the expressions for $f_{	ext{fric}}$ and $d\vec{s}$, we get

$$\vec{f}_{	ext{fric}}\cdot d\vec{s} = -\mu mg\cos	heta r\sin	heta d	heta = -\mu mg(1-\cos^2	heta)r d	heta$$

Integrating this expression from $	heta_A = 0$ to $	heta_B$, we get

$$W_{	ext{fric}} = \int_0^{	heta_B} -\mu mg(1-\cos^2	heta)r d	heta = -\mu mg r (	heta_B - \frac{1}{2}\sin 2	heta_B)$$

Plugging in the given values, we have

$$\Delta E = \frac{1}{2}(1.2	ext{ kg})(5.6	ext{ m/s})^2 - \frac{1}{2}(1.2	ext{ kg})(8.4	ext{ m/s})^2 - (-0.80	ext{ m})(0.50)(1.2	ext{ kg})(9.81	ext{ m/s}^2)(	heta_B - \frac{1}{2}\sin 2	heta_B)$$

where we have used $\mu = 0.50$ for the coefficient of kinetic friction. Solving for $	heta_B$ using a numerical method, we find

$$	heta_B \approx 1.03	ext{ rad}$$

Plugging this back in, we get

$$\Delta E \approx -4.7	ext{ J}$$

Therefore, the answer is (C), with a change in mechanical energy of $-4.7$ J.

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