Question

A 16kg sled starts up a 28 degree incline with a speed of 2.4m/s . The coefficient of kinetic friction is = 0.27.
part a)How far up the incline does the sled travel?

part b)What condition must you put on the coefficient of static friction if the sled is not to get stuck at the point determined in part (a)?

please if you can show all work thank you

Answer 1

(a) Initial kinetic energy =
(Potential energy increase) + (work against friction)
Let X be the distance traveled up the incline.
(M/2)V^2 = M*g*X sin28 + M*g*cos28*(0.27)X

M cancels out. Solve for X

X = (V^2/2g)/[sin28 + 0.27cos28]

(b) To make sure it slides back down the incline,the downward weight component must exceed the maximum possible static friction force.
Us*cos28*M*g < M*g*sin28
Us < tan28
(Us is the static friction coefficient)

Answer 2

i got 65 for part a, but it says its not right.

Answer 3

KE =m•v²/2,
PE =m•g•h = m•g •s•sinα,
W(fr) = F(fr) •s = k•m•g•cosα •s

KE = PE +W(fr)
v²/2 = g •s(sinα + k•cosα),
s = v²/2•g •(sinα + k•cosα)=
=0.415 m.