Question

A boy holds a 40-N weight at arm's length for 10 seconds. His arm s 1.5 m above the ground. What is the work done by the force of the boy on the weight while he is holding?

Answer 1

`60` joules

Explanation:

Work done is given by the equation:

`W=F*d`

where:

• `W` is the work done in joules

• `F` is the force applied in newtons

• `d` is the distance moved in meters

So, we get:

`W=40 \ "N"*1.5 \ "m"`

`=60 \ "J"`

Answer 2

0 J

Explanation:

Work done is given by the equation:

`W = vecF*vecd*costheta`
where `theta` is the angle between `vecF and vecd`.

`vecF` is the force applied and `vecd` is the displacement of the object the force is applied to. In this case, the object is not moved, so `vecd = 0` and therefore the work done is zero.

A bit of discussion about a slight variation to the situation:
What if the boy walked across the room with the weight held in the position described?
The work done would still be zero. `vecd` would no longer be zero, but the vectors `vecF and vecd` are in perpendicular directions.

Therefore `theta = 90^@ and cos90^@ = 0`.

I hope this helps,
Steve