Question

A box is released from rest at the top of a ramp. The surface of the ramp makes an angle of 31.5degrees with the horizontal, and is rough; the coefficients of friction between the ramp and the box are us = 0.500 and pk = 0.350. What is the magnitude of the acceleration of the box?

Answer 1

7N mgsin31.5 mycos31.5 mg 31.5 deg

Consider the figure

Weight of the box, is acting vertically downwards, split it into its components.

би

is the frictional force

f

is the normal force

Frictional Force = Coefficient of friction*Normal force

=============

Consider the box

Net Force = Mass*Acceleration

$mgsin31.5-f=ma$

$mgsin31.5-\mu _{k}N=ma$

$mgsin31.5-\mu _{k}mgcos31.5=ma$

Cancel mass m

$gsin31.5-\mu _{k}gcos31.5=a$

$9.81m/s^{2}*sin31.5-0.350*9.81m/s^{2}*cos31.5=a$

$9.81*sin31.5-0.350*9.81*cos31.5=a$

$5.12571-2.92754=a$

ANSWER: ${\color{Red} a=2.19817m/s^{2}}$

=============================