Question

A 9.00 meter long ramp is inclined at an angle of 6.00 above horizontal. A 7 kg box is given an intial push at the bottom of the ramp so that when it gets to the top it stops. The coefficient of kinetic friction between the box and the ramp is .230

a) what is the initial speed of the box just as it is pushed. (The answer is 7.68 m/s, but I can't seem to set it up properly to get 7.68 as an answer)

b) Suppose the box slides back down the ramp, what is the maximum value of its coefficient of static friction? Does that value make sense? Why or why not?

Answer 1

(a) given mass of box is m=7 kg

the weight of box is mg= 7X9.8=68.6 N

Normal force F_N = mgcos6 = 68.22 N

force due friction F_f = f_k F_N = 15.7 N

component of force down the ramp F_d = mg sin6 = 7.17 N

Now, the height at which the box can be pushed is h= Lsin6= 9Xsin6= 0.94 m

and the potential energy at this height is mgh= 64.5 J

and the kinetic energy at bottom is thedifference between loss in potential energy and work done by friction.

$\therefore$ KE= mgh - F_fx L = 64.8 - 15.7 x 9 = 76.5 J

and KE = (1/2)mv² or v = (KE x 2/m)^1/2 = 4.68 m/s

(b) static friction when the box slides back, net force = downward force - force applied due to static friction

since there is no acceleration in static case and net force will be zero. and downward force is equal to force applied by static friction.

$\therefore$ mg sin 6 = f_s mg cos6, or f_s = 0.105