A cable exerts a constant upward tension of magnitude 1.50 ✕ 104 N on a 1.20 ✕ 103 kg elevator as it rises through a vertical distance of 4.30 m.
(a) Find the work done by the tension force on the elevator (in J).

(b) Find the work done by the force of gravity on the elevator (in J).

Let's solve each part step-by-step.

### Given dataLet's solve each part step-by-step.

### Given 
- Tension force \( T = 1.50 \times 10^4 \, \text{N} \)
- Mass of the elevator \( m = 1.20 \times 10^3 \, \text{kg} \)
- Distance \( d = 4.30 \, \text{m} \)
- Gravitational acceleration \( g = 9.81 \, \text{m/s}^2 \)

---

### Part (a): Work done by the tension force

The work done by a force acting in the direction of motion is calculated as:
\[
W = F \cdot d
\]

Since the tension force acts upward and the elevator is moving upward, they are in the same direction, so:
\[
W_{\text{tension}} = T \cdot d = (1.50 \times 10^4 \, \text{N}) \cdot (4.30 \, \text{m})
\]

Calculating this:
\[
W_{\text{tension}} = 1.50 \times 10^4 \times 4.30 = 64500 \, \text{J}
\]

So, the work done by the tension force is:
\[
\boxed{64500 \, \text{J}}
\]

---

### Part (b): Work done by the force of gravity

The force of gravity on the elevator is given by:
\[
F_{\text{gravity}} = m \cdot g = (1.20 \times 10^3 \, \text{kg}) \cdot (9.81 \, \text{m/s}^2)
\]

Calculating this:
\[
F_{\text{gravity}} = 1.20 \times 10^3 \times 9.81 = 11772 \, \text{N}
\]

Since the gravitational force acts downward and the elevator is moving upward, the angle between the force of gravity and the displacement is \( 180^\circ \), resulting in a negative work done by gravity.

Thus, the work done by gravity is:
\[
W_{\text{gravity}} = -F_{\text{gravity}} \cdot d = -(11772 \, \text{N}) \cdot (4.30 \, \text{m})
\]

Calculating this:
\[
W_{\text{gravity}} = -11772 \times 4.30 = -50619.6 \, \text{J}
\]

So, the work done by the force of gravity is:
\[
\boxed{-50619.6 \, \text{J}}
\]

---

### Summary of answers:
- (a) Work done by the tension force: \( \boxed{64500 \, \text{J}} \)
- (b) Work done by the force of gravity: \( \boxed{-50619.6 \, \text{J}} \)

Would you like further details on this solution, or do you have any other questions?

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1. What is the net work done on the elevator, considering both tension and gravity?
2. How would the answers change if the elevator were moving downward instead?
3. What would happen to the work done if the tension force were less than the gravitational force?
4. How is work related to energy in this scenario?
5. How would friction affect the work done by tension?

**Tip:** Always pay attention to the direction of forces when calculating work, as it affects whether work is positive or negative.

A cable exerts a constant upward tension of magnitude 1.50 ✕ 10^4 N on a 1.20 ✕ 10^3 kg elevator as it rises through a vertical distance of 4.30 m. (a) Find the work done by the tension force on the elevator (in J). (b) Find the work done by the force of gravity on the elevator (in J).

Learn how to calculate the work done by the tension force and gravity on an elevator as it rises 4.30 meters. This problem covers key concepts in physics, such as forces, work-energy theorem, and gravitational force. Suitable for students in Grades 10-12.

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