Question

practice: work calculations
answer the following questions. make sure to show all work to receive credit. you may need a separate sheet of paper.

1. it took 500 n of force to move a car 4 meters. how much work was done?
2. a force of 15 n is used to push a box along the floor a distance of 1.2 meters. find the amount of work done.
3. a force of 110 n was needed to lift a rock. a total of 150 j of work was done. how far was the rock lifted?
4. max exerted a force of 9,000 n on a stalled car but was unable to move it. how much work did he do?
5. a 5.0 kg rock is lifted 2.0 m. find the amount of work done.

Explanation:

Step1: Recall the work formula

The formula for work is \(W = F\times d\), where \(W\) is work, \(F\) is force, and \(d\) is distance.

Step2: Substitute values for question 1

Given \(F = 500\space N\) and \(d=4\space m\). Substitute into the formula: \(W = 500\times4\).

Step3: Calculate for question 1

\(W = 2000\space J\).

Step4: Substitute values for question 2

Given \(F = 15\space N\) and \(d = 1.2\space m\). Substitute into \(W=F\times d\): \(W=15\times1.2\).

Step5: Calculate for question 2

\(W = 18\space J\).

Step6: Rearrange formula for question 3

From \(W = F\times d\), we get \(d=\frac{W}{F}\). Given \(W = 150\space J\) and \(F = 110\space N\). Substitute: \(d=\frac{150}{110}\).

Step7: Calculate for question 3

\(d\approx1.36\space m\).

Step8: Analyze question 4

Since the car is not moved (\(d = 0\)), using \(W=F\times d\), \(W=9000\times0\).

Step9: Calculate for question 4

\(W = 0\space J\).

Step10: Find force for question 5

First, find the force \(F=mg\) (where \(m = 5.0\space kg\) and \(g = 9.8\space m/s^{2}\)). So \(F=5.0\times9.8=49\space N\). Then use \(W=F\times d\) with \(d = 2.0\space m\). Substitute: \(W=49\times2.0\).

Step11: Calculate for question 5

\(W = 98\space J\).

Answer:

1. \(2000\space J\)
2. \(18\space J\)
3. \(\approx1.36\space m\)
4. \(0\space J\)
5. \(98\space J\)