Question

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?


It is less than 2.5 N.

It is greater than 2.5 N.

It equals 2.5 N.

Answer 1

To determine the relationship of the output force with respect to the given work input and output distance while considering the effect of friction, we can use the work-energy principle.

The work done (input work) can be expressed as:

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

In this case, the work input is 25 J.

The output work done can also be expressed as:

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

Given that the output distance is 10 m, we can derive the output force.

First, we find the maximum theoretical output force assuming no friction:

\[
\text{Output Work (maximum)} = \text{Input Work} = 25 \, \text{J}
\]

Assuming no friction, then:

\[
\text{Output Force} = \frac{\text{Output Work}}{\text{Output Distance}} = \frac{25 \, \text{J}}{10 \, \text{m}} = 2.5 \, \text{N}
\]

However, since friction is present, the output work (and thus the output force) will be less than this theoretical value. Therefore, the actual output force must be less than 2.5 N.

Thus, considering the effect of friction, it must be true that:

**Output force is less than 2.5 N.**