Answer to Question #100597 in Mechanics | Relativity for Micah Smith

Question

A 77.0 kg box slides down a 25.0◦
ramp with
an acceleration of 3.30 m/s2
.Find µk between the box and the ramp.
The acceleration of gravity is 9.81 m/s
2
.

Answer 1

From the equation $F=ma$

$mgSin(25)-F' =ma \to(1)\newline$

Where m is the mass of the box ,F' is the frictional force acted on the box and 'a' is the acceleration

$\because F'=77.0*9.81*sin(25)-77.0*3.30 \newline F'=65.13N$

Considering the Forces perpendicular to the ramp

$R=mgCos(25)$

solving this,

$R=77.0*9.81Cos(25) \newline R=684.60N$

Therefore the Frictional force acting on the box,

$F'=\mu R \newline \mu=\frac{F'}{R} \newline \mu=0.095$