Question

1. suppose work input is 25 j, and the output distance is 10 m. factoring in the effect of friction, which must be true about output force? (1 point)

○ it is greater than 2.5 n.
○ it is less than 2.5 n
○ it equals 2.5 n.

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: Calculate ideal output force (without friction)

If there were no friction, the work input would equal the work output (\( W_{input} = W_{output} \)). Given \( W_{input} = 25 \, \text{J} \) and \( d_{output} = 10 \, \text{m} \), we can solve for the ideal output force \( F_{ideal} \):
\( F_{ideal} = \frac{W_{output}}{d_{output}} = \frac{25 \, \text{J}}{10 \, \text{m}} = 2.5 \, \text{N} \).

Step3: Analyze the effect of friction

Friction converts some of the input work into heat (or other forms of energy), so the actual work output (\( W_{actual} \)) is less than the work input (\( W_{input} \)). Using \( W_{actual} = F_{actual} \times d_{output} \), since \( W_{actual} < W_{input} \) and \( d_{output} \) remains \( 10 \, \text{m} \), we have:
\( F_{actual} = \frac{W_{actual}}{d_{output}} < \frac{W_{input}}{d_{output}} = 2.5 \, \text{N} \).

Answer:

It is less than 2.5 N.