Asked by korede
A driver of an 1800kg/hr including passengers travelling at 25m/s slams the brake locking the wheels on the dry pavement. The coefficient of kinetic friction between the rubber and the dry concrete is typically 0.7.calculate how far the car will travel before stopping.
Answers
Answered by
korede
answer is 38.6 but I can figure out how
how do I get frictional force and is the normal reaction mg or 0
how do I get frictional force and is the normal reaction mg or 0
Answered by
Damon
1800 kg * 9.81 = 17,658 Newtons weight
friction force = .7 * 17,658 = 12,360 Newtons retarding force
acceleration = - force/mass = -12360/1800 = - 6.87 m/s^2
note - we could have said .7 g = 6.87
v = 25 - 6.87 t
v=0 when t = 3.64 seconds
x = Xi + Vi t - (6.87/2) t^2
= 0 + 25*3.64 - 3.44 *3.64^2
= 91 - 45.6 = 45.5 meters
BUT more easily the average speed during stop is 25/2 = 12.5 m/s
12.5 * 3.64 = 45.5 meters
friction force = .7 * 17,658 = 12,360 Newtons retarding force
acceleration = - force/mass = -12360/1800 = - 6.87 m/s^2
note - we could have said .7 g = 6.87
v = 25 - 6.87 t
v=0 when t = 3.64 seconds
x = Xi + Vi t - (6.87/2) t^2
= 0 + 25*3.64 - 3.44 *3.64^2
= 91 - 45.6 = 45.5 meters
BUT more easily the average speed during stop is 25/2 = 12.5 m/s
12.5 * 3.64 = 45.5 meters
Answered by
Damon
By the way, no way 38.6 meters (on earth anyway).
Answered by
henry2,
M*g = 1800 * 9.8 = 17,640 N. = Wt. of vehicle = Normal force, Fn.
u*Fn = 0.7 * 17,640 = 12,348 N. = Force of kinetic friction.
-uFn = M * a.
-12,348 = 1800 * a.
a = -686 m/s^2.
V^2 = Vo^2 + 2a*d = 0.
25^2 + (-13.72)d = 0,
d = 45.6 m.
u*Fn = 0.7 * 17,640 = 12,348 N. = Force of kinetic friction.
-uFn = M * a.
-12,348 = 1800 * a.
a = -686 m/s^2.
V^2 = Vo^2 + 2a*d = 0.
25^2 + (-13.72)d = 0,
d = 45.6 m.
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