Solution to Question 47
Physics Principle: Work-Energy Theorem for Coupled Forces.
Given Parameters:
\(h = 1.25 \text{ m}\) (Height)
\(v_0 = 3.0 \text{ m/s}\) (Velocity in Case 1)
\(\mu = 0.375\) (Friction coefficient)
Angle \(\theta = 37^\circ\) (from 3-4-5 triangle in diagram: \(\sin 37^\circ = 0.6, \cos 37^\circ = 0.8, \cot 37^\circ = 1.33\)).
Step 1: Formulate Work-Energy Equations
We consider two cases. In both cases, the block moves down the incline.
Case 1 (Attraction): Charges have opposite signs. The electric force \(F_e\) has a component perpendicular to the surface that increases or decreases the normal force, and a component along the motion that does work. Let \(W_e\) be the work done by the conservative electric force. Let \(W_{fric}^{(1)}\) be the work done by friction.
$$ W_{grav} + W_e + W_{fric}^{(1)} = \frac{1}{2}m v_0^2 $$
The normal force is \(N_1 = mg \cos\theta + F_{e,\perp}\) (Assuming attraction adds to the normal component towards O).
$$ W_{fric}^{(1)} = -\mu \int N_1 dl = -\mu \int (mg \cos\theta + F_{e,\perp}) dl $$
$$ W_{fric}^{(1)} = -\mu W_{N,grav} – \mu \int F_{e,\perp} dl $$
Case 2 (Repulsion): Block charge is reversed. Electric force direction reverses \(\vec{F} \to -\vec{F}\).
Work by electric force becomes \(-W_e\).
Normal force becomes \(N_2 = mg \cos\theta – F_{e,\perp}\).
$$ W_{grav} – W_e + W_{fric}^{(2)} = \frac{1}{2}m v^2 $$
$$ W_{fric}^{(2)} = -\mu \int (mg \cos\theta – F_{e,\perp}) dl = -\mu W_{N,grav} + \mu \int F_{e,\perp} dl $$
Step 2: Combine the Equations
Let’s add the energy equations for Case 1 and Case 2:
Eq 1: \( W_g + W_e – \mu W_{N,grav} – \mu \int F_{e,\perp} dl = K_1 \)
Eq 2: \( W_g – W_e – \mu W_{N,grav} + \mu \int F_{e,\perp} dl = K_2 \)
Summing them cancels out all terms involving the electric force (both the direct work \(W_e\) and the frictional modification):
$$ 2 W_g – 2 \mu W_{N,grav} = K_1 + K_2 $$
$$ 2 (mgh) – 2 \mu (mg \cos\theta \cdot L) = \frac{1}{2}m v_0^2 + \frac{1}{2}m v^2 $$
where \(L\) is the path length, \(L = h / \sin\theta\).
Substitute \(L\): $$ 2mgh – 2\mu mg \cos\theta \frac{h}{\sin\theta} = \frac{1}{2}m(v_0^2 + v^2) $$ $$ 2gh (1 – \mu \cot\theta) = \frac{1}{2}(v_0^2 + v^2) $$ $$ 4gh (1 – \mu \cot\theta) = v_0^2 + v^2 $$ $$ v = \sqrt{ 4gh(1 – \mu \cot\theta) – v_0^2 } $$
Step 3: Calculation
\( \mu \cot\theta = 0.375 \times \frac{4}{3} = \frac{3}{8} \times \frac{4}{3} = \frac{1}{2} = 0.5 \)
\( 1 – \mu \cot\theta = 0.5 \)
\( 4gh (0.5) = 2gh = 2 \times 10 \times 1.25 = 25 \)
Given \( v_0 = 3 \implies v_0^2 = 9 \)
$$ v = \sqrt{ 25 – 9 } = \sqrt{16} = 4.0 \text{ m/s} $$