# What is the work done by a 0.47 N force pushing a 0.026 kg pencil 0.26 m against a force of friction of 0.23N?

Dec 23, 2017

It depends, between $\text{0.06" - "0.12 J}$.

#### Explanation:

Work $= F d \cos \theta$

• Meaning the work is done effectively only by the component of the force that is in (or opposite to) the direction traveled by the object when it is being pushed or pulled.

Hence, it depends on the direction, which is not specified, of the force being applied.

If the force is applied directly opposite to the friction, then work done by the force is

(a) $\text{ "W = Fd = "0.47 N" * ".26 m" = "0.12 Nm" = "0.12 J}$

(Note: keep the answer in 2 significant digits only because that is how the force and distance are given, in 2 significant digits.)

However, if the force is directed at an angle, then its component $F \cos \theta$ must be at least as much as the friction, otherwise, the pencil won't move. In this case, the force is effectively exerting only $F \cos \theta = \text{0.23 N}$ on the pencil. Thus,

(b) $\text{ "W =(Fcos theta) d = "0.23 N" *".26 m" = "0.060 J}$

Note that in Case (a), the force is bigger than the friction, hence the pencil actually accelerates, meaning part of the work done is converted into kinetic energy of the pencil. In Case (b), there is no acceleration, i.e., constant speed, the pencil's kinetic energy does not change.