All Work and No Play

Learning Goal: To be able to calculate work done by a constant force directed at different angles relative to displacement

If an object undergoes displacement while being acted upon by a force (or several forces), it is said that work is being done on the object. If the object is moving in a straight line and the displacement and the force are known, the work done by the force can be calculated as

 

where W is the work done by forceon the object that undergoes displacement directed at angle θ relative to .

Note that depending on the value of  cosθ , the work done can be positive, negative, or zero.

In this problem, you will practice calculating work done on an object moving in a straight line. The first series of questions is related to the accompanying figure.

A) The work done by force is

=> It is zero

 

B) The work done by force is

=> It is positive

 

C) The work done by force is

=> negative

 

D) The work done by force is

=> positive

 

E) The work done by force is

=> negative

 

F) The work done by force is

=> zero

 

G) The work done by force is

=> positive

H) Find the work W done by the 18-newton force.

=> W= 2900 J

 

I) Find the work W done by the 30-newton force.

=> W= 4200 J

 

J) Find the work W done by the 12-newton force.

=> W= -1900 J

 

K) Find the work W done by the 15-newton force.

W= -1800 J


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Comments

2 responses to “All Work and No Play”

  1. Erica Avatar

    How do you get the answers

    1. Lyssi Avatar
      Lyssi

      For the first part:
      It’s just cos(theta). If theta is 90 or 270 degrees, the sign of work would be zero because cos90=0 and cos270=0. If the horizontal work is to the right, the sign would be positive. If the horizontal work is to the left, the sign would be negative.

      For the sencond part:
      W=F*s*cos(theta)
      W=Force*distance*cos(theta)
      so J)
      W=12N*160m*cos(180)
      W=-1900J

      Hope this helps!

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