## 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

## Comments ( 2 )

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!