To calculate how far the box travels up the incline before coming to rest, we can use the principles of energy and forces acting on the box. We'll consider the forces acting on the box, the work done against friction, and the change in kinetic energy.
Step 1: Identify the forces acting on the box
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Gravitational force (mg) acting downwards, which can be decomposed into two components:
- \( mg \sin(\theta) \) acting down the incline.
- \( mg \cos(\theta) \) acting perpendicular to the incline.
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Frictional force (\( f_k \)) opposing the motion, which can be described by: \[ f_k = \mu_k \cdot N = \mu_k \cdot (mg \cos(\theta)) \] where \( \mu_k = 0.2 \) is the coefficient of kinetic friction.
Step 2: Calculate the components of the forces
- Let \( m \) be the mass of the box and \( g \approx 9.81 , \text{m/s}^2 \).
- It makes an angle \( \theta = 20^\circ \).
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Gravitational force components: \[ F_{\text{gravity}} = mg \sin(20^\circ) \] \[ F_{\text{normal}} = mg \cos(20^\circ) \]
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Frictional force: \[ f_k = 0.2 \cdot (mg \cos(20^\circ)) \]
Step 3: Calculate the effective force acting on the box
The net force acting on the box as it slides up is: \[ F_{\text{net}} = -F_{\text{gravity}} - f_k \]
Substituting the expressions: \[ F_{\text{net}} = -mg \sin(20^\circ) - 0.2 \cdot (mg \cos(20^\circ)) \] \[ F_{\text{net}} = -mg \left( \sin(20^\circ) + 0.2 \cos(20^\circ) \right) \]
Step 4: Use initial kinetic energy to find the stopping distance
The initial kinetic energy (KE) of the box is: \[ KE = \frac{1}{2} mv^2 = \frac{1}{2} m (2 , \text{m/s})^2 = 2m , \text{J} \]
Using work-energy principle, the work done against the net force will be equal to the initial kinetic energy: \[
- \text{Work} = F_{\text{net}} \cdot d = KE \] Thus, \[ mg \left( \sin(20^\circ) + 0.2 \cos(20^\circ) \right) \cdot d = 2m \]
Step 5: Solve for distance \( d \)
The mass \( m \) cancels out from both sides: \[ g \left( \sin(20^\circ) + 0.2 \cos(20^\circ) \right) \cdot d = 2 \]
Now substituting \( g \approx 9.81 , \text{m/s}^2 \): \[ d = \frac{2}{9.81 \times \left( \sin(20^\circ) + 0.2 \cos(20^\circ) \right)} \]
Calculating:
- \( \sin(20^\circ) \approx 0.342 \)
- \( \cos(20^\circ) \approx 0.940 \)
Thus, \[ \sin(20^\circ) + 0.2 \cos(20^\circ) \approx 0.342 + 0.2 \times 0.940 = 0.342 + 0.188 = 0.530 \]
Now substituting back: \[ d = \frac{2}{9.81 \times 0.530} = \frac{2}{5.1973} \approx 0.384 \text{ meters} \]
Conclusion
The box travels approximately 0.384 meters along the incline before coming to rest.