A forklift exerts an upward force of 2.00 × 10³ N on a box as X it moves the box 5.00 m higher. How much work does the forklift do?

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Chapter5: Energy
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Problem 4P: a shopper in a supermarket pushes a cart with a force of 35 N directed at an angle of 25 below the...
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### Work Calculation Problem

**You may use a calculator.**

---

**Problem:**

A forklift exerts an upward force of \( 2.00 \times 10^3 \) N on a box as it moves the box 5.00 m higher. How much work does the forklift do?

---

**Answer Choices:**

- \( \circ \ 1.00 \times 10^4 \) J
- \( \circ \ 500 \) J
- \( \circ \ 0.00 \) J
- \( \circ \ 1.00 \times 10^3 \) J

**Selected Answer:**
- \( \bullet \ 0.00 \) J

---

**Explanation:**

In this problem, we are asked to calculate the work done by a forklift. Work can be calculated using the formula:

\[ \text{Work} = \text{Force} \times \text{Distance} \times \cos(\theta) \]

Where:
- Force (\(F\)) = \( 2.00 \times 10^3 \) N
- Distance (\(d\)) = 5.00 m
- \( \theta \) is the angle between the force and the direction of movement. In this case, since the force exerted by the forklift is vertically upward and the distance moved is also vertically upward, \( \theta \) = 0 degrees. Therefore, \( \cos (0) \) = 1.

Thus, the work done:

\[ \text{Work} = (2.00 \times 10^3 \text{ N}) \times (5.00 \text{ m}) \times 1 = 10.00 \times 10^3 \text{ J} \]
\[ \text{Work} = 1.00 \times 10^4 \text{ J} \]

The amount of work done by the forklift is therefore \( 1.00 \times 10^4 \) J. Hence, the selected answer (0.00 J) is not correct.
Transcribed Image Text:### Work Calculation Problem **You may use a calculator.** --- **Problem:** A forklift exerts an upward force of \( 2.00 \times 10^3 \) N on a box as it moves the box 5.00 m higher. How much work does the forklift do? --- **Answer Choices:** - \( \circ \ 1.00 \times 10^4 \) J - \( \circ \ 500 \) J - \( \circ \ 0.00 \) J - \( \circ \ 1.00 \times 10^3 \) J **Selected Answer:** - \( \bullet \ 0.00 \) J --- **Explanation:** In this problem, we are asked to calculate the work done by a forklift. Work can be calculated using the formula: \[ \text{Work} = \text{Force} \times \text{Distance} \times \cos(\theta) \] Where: - Force (\(F\)) = \( 2.00 \times 10^3 \) N - Distance (\(d\)) = 5.00 m - \( \theta \) is the angle between the force and the direction of movement. In this case, since the force exerted by the forklift is vertically upward and the distance moved is also vertically upward, \( \theta \) = 0 degrees. Therefore, \( \cos (0) \) = 1. Thus, the work done: \[ \text{Work} = (2.00 \times 10^3 \text{ N}) \times (5.00 \text{ m}) \times 1 = 10.00 \times 10^3 \text{ J} \] \[ \text{Work} = 1.00 \times 10^4 \text{ J} \] The amount of work done by the forklift is therefore \( 1.00 \times 10^4 \) J. Hence, the selected answer (0.00 J) is not correct.
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