Fc = mg = 2kg * 9.8N/kg = 19.6N = Force
of car.
Fc = 19.8N @ 0 deg.(Hor. plane).
Fp = mg*sin(0) = 19.8sin(0) = 0 = Force
parallel with plane.
Fv = 19.6cos(0) = 19.6N = Force perpendicular to plane.
Ff = u*Fv = 0.5 * 19.6 = 9.8N.
Fn = Fp - Ff = 0 - 9.8 = -9.8N = Net
force.
Fn = ma,
a = Fn/m = -9.8/2 = -4.9m/s^2.
Vf^2 = Vo^2 + 2ad,
d = (Vf^2-Vo^2) / 2a,
d = (0-(13)^2 / -9.8 = 17.2m. = Length
of skid marks.
A 2000 {\rm kg} car traveling at a speed of 13 {\rm m/s} skids to a halt on wet concrete where mu_k = 0.50. How long are the skid marks?
I tried doing this many times and each answer is wrong. Can someone help?
5 answers
Correction:
mass of the car = 2000kg: NOT 2kg.
Fc = 2000kg * 9.8N/kg = 19,600N.
Fp = 19,600sin(0) = 0.
Fv = 19,600cos(0) = 19,600N.
Ff = u*Fv = 0.5 * 19,600 = 9800N.
Fn = Fp - Ff = o - 9800 = -9800N.
a = Fn/m = -9800 / 2000 = -4.9m/s^2.
The remaining calculations are correct
as is.
mass of the car = 2000kg: NOT 2kg.
Fc = 2000kg * 9.8N/kg = 19,600N.
Fp = 19,600sin(0) = 0.
Fv = 19,600cos(0) = 19,600N.
Ff = u*Fv = 0.5 * 19,600 = 9800N.
Fn = Fp - Ff = o - 9800 = -9800N.
a = Fn/m = -9800 / 2000 = -4.9m/s^2.
The remaining calculations are correct
as is.
Thank you very much!
Glad i could help!
work = change in kinetic energy
friction force X distance = change in KE
-mu_k n x = 0 - 1/2 m v^2
[Remember, friction does negative work)
mu_k m g x = 1/2 m v^2
Simplifying:
x = v^2/(2 mu_k g) = 17.24 m
Notice that in this solution, the mass of the car is not needed.
friction force X distance = change in KE
-mu_k n x = 0 - 1/2 m v^2
[Remember, friction does negative work)
mu_k m g x = 1/2 m v^2
Simplifying:
x = v^2/(2 mu_k g) = 17.24 m
Notice that in this solution, the mass of the car is not needed.
1 answer