# A 10.0\ \mathrm{kg} crate initially at rest sitting on the floor with a static and kinetic...

## Question:

A {eq}10.0\ \mathrm{kg} {/eq} crate initially at rest sitting on the floor with a static and kinetic coefficient of 0.20 is pushed with a force of {eq}15\ \mathrm{N} {/eq} for {eq}4.0\ \mathrm{s} {/eq}. What is the work done?

## Work

The concept of work in physics measures the energy transferred to an object when it is moved some distance by an external force. For a constant external force this can be written as:

{eq}\displaystyle W = F_{\parallel} \Delta x {/eq}

Here:

{eq}\displaystyle W {/eq} is the work done on the object,

{eq}\displaystyle F_{\parallel} {/eq} is component of the external force applied to the object in the direction of the motion, and

{eq}\displaystyle \Delta x {/eq} is the distance the force is applied over.

To find the work done on the crate we first need to determine whether the force acting on the crate results in motion or not. If the applied force is not large enough to overcome static friction then the crate does not move and thus no work is done.

The force of static friction is given by the formula:

{eq}\displaystyle F_f \leq \mu_s F_N {/eq}

Here:

{eq}\displaystyle \mu_s {/eq} is the coefficient of static friction, and

{eq}\displaystyle F_N = mg {/eq} is the normal force of the crate which depends on the crates mass {eq}\displaystyle m {/eq} and the acceleration due to gravity {eq}\displaystyle g \approx 9.8 \: \text{m/s}^2 {/eq}.

Calculating the maximum force of static friction on the crate yields:

{eq}\displaystyle F_f = (0.2)(10.0 \: \text{kg})(9.8 \: \text{m/s}^2) = 19.6 \: \text{N} {/eq}

Since the crate is only pushed with an applied force of 15 N this is insufficient to overcome static friction. Thus the crate does not move at all ({eq}\displaystyle \Delta x = 0 {/eq}) and thus no work is done on the crate. 