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Problem Sets:
Problem
A small object is initially at the bottom of a slope inclined at an angle $\alpha$ with the horizontal. It is projected upward along the slope with an initial velocity $v_0$, and reaches the maximum height. It then slides downward and returns to the initial position with final velocity $v_1$. The coefficient of sliding friction between the object and the slope surface is $\mu$.
- The average speed going up the slope is $v_{up}$. Average speed coming down the slope is $v_{down}$. What is the ratio $v_{up}/v_{down}$?
- What is the ratio of the initial speed before going up the slope and the final speed after coming down from the slope $v_0/v_1$?
Hints
Hint 1:
The ratio of the average speed going up to the average speed coming down is:
$$ \frac{v_{up}}{v_{down}} = \sqrt{\frac{\sin\alpha + \mu\cos\alpha}{\sin\alpha - \mu\cos\alpha}} = \sqrt{\frac{\tan\alpha + \mu}{\tan\alpha - \mu}} $$The ratio of the initial speed before going up to the final speed after coming down is:
$$ \frac{v_0}{v_1} = \sqrt{\frac{\sin\alpha + \mu\cos\alpha}{\sin\alpha - \mu\cos\alpha}} = \sqrt{\frac{\tan\alpha + \mu}{\tan\alpha - \mu}} $$Here is the solution to the problem, following the specified format and guidelines.
We begin by analyzing the forces and determining the object's acceleration during its upward and downward motion. Let the positive direction be up the slope. The acceleration due to gravity is $g$.
1. Acceleration on the Slope
The forces acting on the object parallel to the slope are the component of gravity, $mg\sin\alpha$, and the kinetic friction force, $f_k = \mu N$. The normal force is $N = mg\cos\alpha$.
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Going Up: Both gravity and friction act down the slope, opposing the upward motion. The net force is $F_{net,up} = -mg\sin\alpha - \mu mg\cos\alpha$. The acceleration going up is:
$$a_{up} = \frac{F_{net,up}}{m} = -g(\sin\alpha + \mu\cos\alpha)$$ -
Coming Down: Gravity acts down the slope, while friction acts up the slope, opposing the downward motion. The net force is $F_{net,down} = -mg\sin\alpha + \mu mg\cos\alpha$. The acceleration coming down is:
$$a_{down} = \frac{F_{net,down}}{m} = -g(\sin\alpha - \mu\cos\alpha)$$
For the object to slide down, the gravitational component must exceed friction, so $\sin\alpha > \mu\cos\alpha$, or $\tan\alpha > \mu$.
[P2] Ratio of Initial and Final Speeds ($v_0/v_1$)
We use the time-independent kinematic equation, $v_f^2 = v_i^2 + 2as$, where $s$ is the displacement. Let $L$ be the distance the object travels up the slope.
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Going Up: The initial velocity is $v_i = v_0$, final velocity is $v_f = 0$, and displacement is $s = L$.
$$0^2 = v_0^2 + 2a_{up}L \implies v_0^2 = -2a_{up}L$$ $$v_0^2 = 2g(\sin\alpha + \mu\cos\alpha)L$$ -
Coming Down: The initial velocity is $v_i = 0$, final velocity is $v_f = -v_1$ (speed $v_1$ in the negative direction), and displacement is $s = -L$.
$$(-v_1)^2 = 0^2 + 2a_{down}(-L) \implies v_1^2 = -2a_{down}L$$ $$v_1^2 = 2g(\sin\alpha - \mu\cos\alpha)L$$
Taking the ratio of $v_0^2$ to $v_1^2$ eliminates the unknown distance $L$:
$$\frac{v_0^2}{v_1^2} = \frac{2g(\sin\alpha + \mu\cos\alpha)L}{2g(\sin\alpha - \mu\cos\alpha)L} = \frac{\sin\alpha + \mu\cos\alpha}{\sin\alpha - \mu\cos\alpha}$$Taking the square root gives the ratio of the speeds. Dividing numerator and denominator by $\cos\alpha$ yields a more compact form:
$$\frac{v_0}{v_1} = \sqrt{\frac{\tan\alpha + \mu}{\tan\alpha - \mu}}$$[P1] Ratio of Average Speeds ($v_{up}/v_{down}$)
For motion under constant acceleration, the average speed over a distance is the average of the initial and final speeds.
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Going Up: Initial speed is $v_0$, final speed is $0$.
$$v_{up} = \frac{v_0 + 0}{2} = \frac{v_0}{2}$$ -
Coming Down: Initial speed is $0$, final speed is $v_1$.
$$v_{down} = \frac{0 + v_1}{2} = \frac{v_1}{2}$$
The ratio of the average speeds is:
$$\frac{v_{up}}{v_{down}} = \frac{v_0/2}{v_1/2} = \frac{v_0}{v_1}$$This is the same ratio as derived in [P2].