Answer to Question #153572 in Mechanics | Relativity for judith

Question

A small block of mass m slides down a rough plane inclined at an angle of 45° to the horizontal.

If the block starts from rest, and has velocity v after travelling a distance x down the plane, find

an exact expression for the work done against friction, in terms of m, g, x and v.

Answer 1

Let's form the movement equations

first: top of the rough plane

the total energy is equal to potential energy because kinetic energy is 0 on top of a rough plane

$E_t = E_p = mgh = mgxsin\alpha$

second: the down of the plane

the total energy is equal to the sum of kinetic energy and work done by against friction force

because in the bottom of the rough plane the potential energy is 0

$E_t = E_k + A =$ $\frac{mv^2}{2} + A$

here A is work done by against friction force

$mgxsin\alpha = \frac{mv^2}{2} + A$

A = $mgxsin\alpha -$ $\frac{mv^2}{2} = mgxsin45^o - \frac{mv^2}{2} = \frac{\sqrt{2}}{2}mgx - \frac{mv^2}{2}$

Answer: A = $\frac{\sqrt{2}}{2}mgx - \frac{mv^2}{2}$