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1. A 47 N force is used to push a 0.025 kg pencil 0.25 m against a 23 N force of friction. a. Calculate the work done by the applied force on the pencil. b. Calculate the work done by the force of friction during the 0.25 m displacement. c. Calculate the net work (overall work) being done during this process.

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Answer:

Step-by-step explanation:

a. To calculate the work done by the applied force on the pencil, we can use the formula:

Work = Force * Distance * cos(theta)

Given:

Applied force = 47 N

Distance = 0.25 m

Theta (angle between the force and displacement) = 0 degrees (assuming the force is applied in the direction of displacement)

Using the formula, we have:

Work = 47 N * 0.25 m * cos(0 degrees)

Work = 11.75 Joules

Therefore, the work done by the applied force on the pencil is 11.75 Joules.

b. To calculate the work done by the force of friction during the 0.25 m displacement, we can use the formula:

Work = Force * Distance * cos(theta)

Given:

Force of friction = 23 N

Distance = 0.25 m

Theta (angle between the force and displacement) = 180 degrees (assuming the force of friction opposes the displacement)

Using the formula, we have:

Work = 23 N * 0.25 m * cos(180 degrees)

Work = -5.75 Joules

Note: The negative sign indicates that the force of friction does negative work, as it opposes the displacement.

Therefore, the work done by the force of friction during the 0.25 m displacement is -5.75 Joules.

c. The net work (overall work) being done during this process can be calculated by adding the work done by the applied force and the work done by the force of friction:

Net Work = Work by applied force + Work by force of friction

Net Work = 11.75 Joules + (-5.75 Joules)

Net Work = 6 Joules

Therefore, the net work (overall work) being done during this process is 6 Joules.

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