Use the information below the answer the following 3 questions. A 50 kg crate is being dragged across a floor by a force of 225 N at an angle of 40o from the horizontal. The crate is dragged a distance of 5.0 m and the frictional force is 60 N.

Question 2 (2 points)
Question 2 options:
The work done on the crate by the applied force is ___x102 Nm. (Give your answer with the correct number of sign digs and do not include units).
Question 3 (2 points)
Question 3 options:
The work done on the crate by the frictional force is -___x102 Nm. (Give your answer with the correct number of sign digs and do not include units).
Question 4 (2 points)
Question 4 options:
The net work done on the crate is ___x102 Nm. (Give your answer with the correct number of sign digs and do not include units).

Hint: Do not use rounded answers in subsequent calculations

2. 8.62×10² Nm

3. 2.30×10² Nm

4. 6.32×10² Nm

Explanation:

2. Determination of the work done by the applied force.

Force (F) = 225 N

Distance (d) = 5 m

Angle (θ) = 40°

Workdone (Wd) =?

Wd = Fd × Cos θ

Wd = 225 × 5 × Cos 40

Wd = 8.62×10² Nm

3. Determination of the work done by the frictional force.

Frictional Force (Fբ) = 60 N

Distance (d) = 5 m

Angle (θ) = 40°

Workdone (Wd) =?

Wd = Fբd × Cos θ

Wd = 60 × 5 × Cos 40

Wd = 2.30×10² Nm

4. Determination of the net work done.

We'll begin by calculating the net force acting on the crate

Force applied (F) = 225 N

Frictional Force (Fբ) = 60 N

Net force (Fₙ) =?

Fₙ = F – Fբ

Fₙ = 225 – 60

Fₙ = 165 N

Finally, we shall determine the net Workdone. This can be obtained as follow:

Net force (Fₙ) = 165 N

Distance (d) = 5 m

Angle (θ) = 40°

Workdone (Wd) =?

Wd = Fₙd × Cos θ

Wd = 165 × 5 × Cos 40

Wd = 6.32×10² Nm

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i think the answer is a, hope this is right, that would be my best guess.