Question

Now the student moves the box up a ramp

(with the same coefficient of friction) inclined

at 10.1

◦ with the horizontal.

b) If the box starts from rest at the bottom

of the ramp and is pulled at an angle of 26.7

◦

with respect to the incline and with the same

173 N force, what is the acceleration up the

ramp?

Answer in units of m/s

2

.

Answer 1

$m=42.4kg$

$a=9.81 m/s2$

Considering the forces perpendicular to the inclined floor 

$mg*cos(10.1)=R+173*sin(26.7)$

solving above equation

$R=331.766 N$

From, F′=μR Where F' is the frictional force

$F′=66.353 N (1)$

Applying $F=ma$ along the inclined floor

$173*cos(26.7)-F'-mg*sin(10.1)=ma$

From (1)

$173*cos(26.7)-66.353-42.4*9.81*sin(10.1)=42.4a$

Solving above equation

$a=0.3598 m/s2$