Physics: Coefficient of Friction from Stopping Distance ?
Determine the coefficient of friction. Show reasoning / work.
Please help me with this. Thank you!!
- oldschoolLv 72 months agoFavourite answer
mg*h=mg*µd so h = µ*d so µ = h/d
h/d = µ
1/2.4 = 0.417
2/5.1 = 0.392
3/7.5 = 0.4
4/9.8 = 0.41
5/12.7 = 0.394
6/14.7 = 0.41
7/17.8 = 0.393
8/20.1 = 0.398
9/22.6 = 0.398
10/24.9 = 0.402
µ is about 0.4
- Dr. ZorroLv 72 months ago
Jim’s kinematic approach can be shortcut by applying the work-energy theorem:the change in kinetic energy of the block equals work done on the block. The change in kinetic energy is zero (it starts and ends with zero speed) and the work done on it is by gravity (an amount mgh) and by friction (- k m g d). So
m g h + (- k mg d) = 0
d = (1/k) * h
So plotting d versus h should be fitted by a straight line through the origin and the calculated slope is 1/k.
- ?Lv 72 months ago
The PE (mgh) translates to KE (1/2 MV)²
Then it slows down by friction deceleration
y(t) = ½gt² + v₀t + y₀, v=at +v₀ and d=vt + d₀ are the basic formula you need to know.
Standard gravity 'g' is -9.80665 m/s²
and you need to know Force due to friction:
f = k N
f = friction force
k = coefficient of friction
N = normal force (the portion of force pushing down)