This contains a few parts to it and i understand if you can\'t help with all of
ID: 1959610 • Letter: T
Question
This contains a few parts to it and i understand if you can't help with all of it, so if you can't, then the part i need help with the most is part d.A box of mass m = 10 kg is sliding down a ramp of angle ? = 45°. The coefficient of friction between the ramp and
the box is µ1 = 0.80. The box is released from rest in point A. In point B, the box continues its motion but
horizontally and it stops in point C, due to friction. The coefficient of friction between the box and the horizontal
portion is µ2 = 0.25. (NB: The initial speed of the horizontal motion is the same with the speed at the base of the
ramp.)
a) Draw the force diagram
of the box when on the ramp.
b) Draw the force diagram of the box when on the horizontal surface.
c) Use Newton’s 2nd law to calculate the acceleration of the box on the ramp, and then use it to calculate the
length of the ramp (A-B), knowing that it takes the box a time t = 3.0 s to reach point B.
d) Calculate the stopping distance between points B and C.
Explanation / Answer
Greetings...
For part(c):
sin(45)=Fparallel/mg----> Fparallel=69.4N
Cos(45)=Fnormal/mg ----> Fnormal=69.4N
The friction force:
Ff =*Fn = 0.8*69.4= 55.5N
Sum of forced down the ramp=m*a
69.4+55.5=10*a ----> a= 12.5 m/s2
I'll try to solve other part.
Best Regards.