# Using a Force equation to find out how much WORK is done.

## Homework Statement

A 2.0 kg block is acted on by the following force
Force=[(2N/m)x-(5N/m2)x2]{ihat}+[(13N/(1/m2)x2-(6/m)x+7)e-x^2/(2m^2)cos{(2∏)x/3m}]{jhat}

as the object moves horizontally along a surface. In the expression for the force, x indicates the position of the block along the horizontal axis. The coefficient of kinetic friction between the block and the horizontal surface is 0.05. How much work does the given force do on the block as the block moves horizontally along the surface from x=0m to x=2.0m?

N is Newtons
m is Meters

I am not sure...

## The Attempt at a Solution

I have not attempted it because I am not sure where to start at all!!!!

Force=[(2N/m)x-(5N/m2)x2]{ihat}+[(13N/(1/m2)x2-(6/m)x+7)e-x^2/(2m^2)cos{(2∏)x/3m}]{jhat}

What is the function?

That is the force function in component form. Force= (2Newtons/Meter)x....
I am not sure where to even start.

Okay, but where do I start if it is the integral F(x) dx????

gneill
Mentor
A 2.0 kg block is acted on by the following force
Force=[(2N/m)x-(5N/m2)x2]{ihat}+[(13N/(1/m2)x2-(6/m)x+7)e-x^2/(2m^2)cos{(2∏)x/3m}]{jhat}

As written, the way I parse it, your force formula looks like this:
$$F = \left [ 2\frac{N}{m}x-5\frac{N}{m^2}x^2 \right ]\,\hat i + \left[\left ( \frac{13 N}{\frac{1}{m^2}}x^2 - \frac{6}{m}x + 7 \right )e^{-\frac{x^2}{2 m^2}} cos(2 \pi \frac{x}{3 m})\right]\,\hat j$$

The units don't look right for the coefficient of the j term.

I know that the 13N is over the entire 1/m^2x^2-6/mx+7