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A large crate sits on the floor of a warehouse. Paul and Bob apply constant horizontal forces to the crate. The force applied by Paul has magnitude 43.0 N and direction 61.0∘ south of west.

How much work does Paul's force do during a displacement of the crate that is 12.0 m in the direction 22.0∘ east of north?

Respuesta :

MrRoyal

The amount of work done is -512.13 joules

The given parameters are:

  • Force = 43N
  • Direction 61° SW
  • Displacement = 12m, 22° NE

The amount of work done (W) is calculated using:

[tex]W = F\times d\times \cos(\theta)[/tex]

Where [tex]\theta[/tex] is the measure of angle between the force and the direction of the crate.

So, we have:

[tex]\theta = 61 + 22 + 90[/tex]

[tex]\theta = 173[/tex]

Recall that:

[tex]W = F\times d\times \cos(\theta)[/tex]

So, we have:

[tex]W =43.0 \times 12.0 \times \cos(173^o)[/tex]

Evaluate cos(173)

[tex]W =-43.0 \times 12.0 \times 0.9925[/tex]

Evaluate the products

[tex]W =-512.13[/tex]

Hence, the amount of work done is -512.13 joules

Read more about work done at:

https://brainly.com/question/18762601

onyebuchinnaji

The work done by Paul's force during the displacement of the crate is -512.15 J.

The given parameters:

  • Force applied by Paul, F = 43 N, at 61⁰
  • Displacement of the carte, d = 12 at 22⁰

The work done by Paul's force during the displacement of the crate is calculated as follows;

[tex]W = F \times dcos (\theta)\\\\[/tex]

where;

θ is the angle between the force and the displacement

θ = 61⁰ + 90⁰ + 22⁰

θ =  173⁰

[tex]W= 43 \times 12 \times cos(173)\\\\W = -512.15 \ J[/tex]

Thus, the work done by Paul's force during the displacement of the crate is -512.15 J.

Learn more about work done here: https://brainly.com/question/8119756

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