An 8.4 kg crate is pulled 4.6 m up a 30∘ incline by a rope angled 16 ∘ above the incline. The tension in the rope is 140 N and the crate's coefficient of kinetic friction on the incline is 0.25. How much work is done by tension, by gravity, and by the normal force?

To solve this problem, we need to calculate the work done by each force acting on the crate as it moves up the incline. The forces involved are tension, gravity, and the normal force.

### Given Data
- Mass of crate, \( m = 8.4 \, \text{kg} \)
- Distance moved along the incline, \( d = 4.6 \, \text{m} \)
- Angle of incline, \( \theta = 30^\circ \)
- Angle of rope with respect to the incline, \( \phi = 16^\circ \)
- Tension in the rope, \( T = 140 \, \text{N} \)
- Coefficient of kinetic friction, \( \mu_k = 0.25 \)
- Gravitational acceleration, \( g = 9.8 \, \text{m/s}^2 \)

### Step 1: Calculate the Work Done by Tension
The work done by a force \( F \) at an angle \( \alpha \) to the displacement \( d \) is given by:
\[
W = F \cdot d \cdot \cos(\alpha)
\]

Since the tension is applied at an angle of \( 16^\circ \) above the incline, we can calculate the work done by the tension force as:
\[
W_{\text{tension}} = T \cdot d \cdot \cos(\phi)
\]
Substituting the values:
\[
W_{\text{tension}} = 140 \times 4.6 \times \cos(16^\circ)
\]
Calculating \( \cos(16^\circ) \approx 0.9613 \):
\[
W_{\text{tension}} = 140 \times 4.6 \times 0.9613 \approx 618.2 \, \text{J}
\]

### Step 2: Calculate the Work Done by Gravity
The component of the gravitational force along the incline is:
\[
F_{\text{gravity, parallel}} = m \cdot g \cdot \sin(\theta)
\]
The work done by gravity is then:
\[
W_{\text{gravity}} = -F_{\text{gravity, parallel}} \cdot d
\]
(The work is negative because gravity opposes the motion up the incline.)

Substitute the values:
\[
F_{\text{gravity, parallel}} = 8.4 \times 9.8 \times \sin(30^\circ)
\]
Since \( \sin(30^\circ) = 0.5 \):
\[
F_{\text{gravity, parallel}} = 8.4 \times 9.8 \times 0.5 = 41.16 \, \text{N}
\]
Then,
\[
W_{\text{gravity}} = -41.16 \times 4.6 \approx -189.34 \, \text{J}
\]

### Step 3: Calculate the Work Done by the Normal Force
The normal force acts perpendicular to the direction of displacement along the incline. Since the normal force has no component along the displacement, the work done by the normal force is:
\[
W_{\text{normal}} = 0 \, \text{J}
\]

### Final Answer
- **Work done by tension**: \( W_{\text{tension}} \approx 618.2 \, \text{J} \)
- **Work done by gravity**: \( W_{\text{gravity}} \approx -189.34 \, \text{J} \)
- **Work done by the normal force**: \( W_{\text{normal}} = 0 \, \text{J} \)

Would you like further clarification on any of these calculations?

Here are some related questions you might find interesting:
1. How would the work done by tension change if the angle of the rope was different?
2. What is the effect of increasing the coefficient of kinetic friction on the work done by friction?
3. How would we calculate the total work done on the crate?
4. How would the work change if the incline angle was steeper?
5. How does the normal force change with different incline angles?

**Tip:** When calculating work done by forces, always check the direction of each force relative to the displacement to determine if the work is positive or negative.

An 8.4 kg crate is pulled 4.6 m up a 30° incline by a rope angled 16° above the incline. The tension in the rope is 140 N and the crate's coefficient of kinetic friction on the incline is 0.25. How much work is done by tension, by gravity, and by the normal force?

Detailed solution to finding the work done by tension, gravity, and normal force on a 8.4 kg crate moving up a 30° incline. Learn key concepts in work and energy, forces on an incline, and trigonometry. Suitable for Grades 10-12 physics students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation