(adsbygoogle = window.adsbygoogle || []).push({}); The froce of friction on a body (moving along a surface at constant speed) can be approximated by [itex] \vec{F} = -F \hat{v}[/itex], where F is a positive constant and [itex] \hat{v} = \frac{\vec{v}}{|\vec{v}|} [/itex] is the unit vector in teh direction of [itex] \vec{v} = \dot{\vec{r}} [/itex] (the velocity of the body)

Prove that hte work done in moving the body from r to r+dr (along a given path) ie dW = F . dr= -F ds in this case where

[itex] ds^2 = d\vec{r}^2 = dx^2 + dy^2 + dz^2 [/itex] where ds iteh element of arc length along the path of the body

Hint: [tex] \frac{\frac{d\vec{r}}{dt}}{|\frac{d\vec{r}}{dt}|} = \frac{d\vec{r}}{|d\vec{r}|} [/tex] and [tex] |d\vec{r}| = \sqrt{d\vec{r} \bullet d\vec{r}} = \sqrt{d\vec{r}^2} = ds [/tex]

ok since v hat can be written like the hint and the denominator can be written like ds so

[tex] \vec{F} \bullet d\vec{r} = -F \frac{d\vec{r}}{ds} \bullet d\vec{r} [/tex]

since dr dot dr is ismply dr^2 which is ds^2 we get what we want.

Is this fine?

If the body is moved along a level floor (in which we take to be the xy plane) from the point (0,0) to (1,1) along two different paths

(i) The straight line through (0,0) and (1,1)

(ii) along teh parabola y = x^2

Calculate the work done in teh two cases and compare the results

Im wondering if they mean the work done by the force of friction of the work done by the person pushing the box

If it was the work done by the friction could we even calculate it without any time given?

Please help! Thanks in advance!

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Calculating Work

**Physics Forums | Science Articles, Homework Help, Discussion**