To solve the problem, we can break it down into parts:
Step 1: Work Done by the Force of Gravity
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Determine the height (h) of the incline: If we denote the length of the incline as \( L \), then the height \( h \) can be found using trigonometry: \[ h = L \cdot \sin(\theta) = L \cdot \sin(30^\circ) = L \cdot \frac{1}{2} = \frac{L}{2} \]
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Calculate the force of gravity acting on the block: The weight \( W \) of the block is given by: \[ W = m \cdot g = 5 , \text{kg} \cdot 9.81 , \text{m/s}^2 = 49.05 , \text{N} \]
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Work done by the force of gravity (Wg): The work done by gravity is given by: \[ W_g = W \cdot h = W \cdot \left(\frac{L}{2}\right) = 49.05 \cdot \frac{L}{2} \]
Step 2: Work Done by Friction Force
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Determine the normal force (N): The normal force on an incline is given by: \[ N = m \cdot g \cdot \cos(\theta) = 5 \cdot 9.81 \cdot \cos(30^\circ) = 5 \cdot 9.81 \cdot \frac{\sqrt{3}}{2} \approx 42.44 , \text{N} \]
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Calculate the frictional force (F_f): The frictional force can be calculated using the coefficient of kinetic friction (\( \mu_k = 0.40 \)): \[ F_f = \mu_k \cdot N = 0.40 \cdot 42.44 \approx 16.976 , \text{N} \]
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Work done by friction (W_f): The work done by friction is negative because it opposes the direction of motion: \[ W_f = -F_f \cdot L = -16.976 \cdot L \]
Step 3: Final Velocity at the Bottom of the Incline
To find the final velocity, we can apply the work-energy principle, which states that the work done on the system is equal to the change in kinetic energy.
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Calculate the total work done (W_total): \[ W_{total} = W_g + W_f = \left(49.05 \cdot \frac{L}{2}\right) - \left(16.976 \cdot L\right) \] Simplifying gives: \[ W_{total} = \left(49.05 \cdot \frac{1}{2} - 16.976\right) \cdot L = \left(24.525 - 16.976\right) \cdot L = 7.549 \cdot L \]
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Set the total work equal to the change in kinetic energy: Since the block starts from rest, the final kinetic energy \( KE \) is: \[ KE = \frac{1}{2} m v^2 \] Setting \( W_{total} \) equal to \( KE \): \[ 7.549 \cdot L = \frac{1}{2} \cdot 5 \cdot v^2 \]
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Solve for \( v \): \[ 7.549 \cdot L = \frac{5}{2} v^2 \implies v^2 = \frac{7.549 \cdot L \cdot 2}{5} = \frac{15.098 \cdot L}{5} = 3.0196 \cdot L \] \[ v = \sqrt{3.0196 \cdot L} \]
Summary:
- Work done by the force of gravity: \( W_g = 49.05 \cdot \frac{L}{2} \)
- Work done by friction force: \( W_f = -16.976 \cdot L \)
- Final velocity at the bottom: \( v = \sqrt{3.0196 \cdot L} \)
To find specific numerical values, you will need the length \( L \) of the incline. Let me know if you have that value!