Asked by MJW
A car can brake to a stop from 60mph in 2.6 seconds. If the car has a weight of 3225 lbs, what force in Newton's acts on the car.
F=ma, vt=v0+at, a=(vt-v0)/t=(0-60mph)/2.6s=-23.1m/s^2
3225 lbs=1462.84 kg
thus F=1462.84kg*(-23.1m/s^2)=-63.3N
F=ma, vt=v0+at, a=(vt-v0)/t=(0-60mph)/2.6s=-23.1m/s^2
3225 lbs=1462.84 kg
thus F=1462.84kg*(-23.1m/s^2)=-63.3N
Answers
Answered by
Henry
Vo = 60mi/h * 1600m/mi * 1h/3600s = 26.67 m/s
V = Vo + a*t = 0
a = -Vo/t = -26.67/2.6 = -10.26 m/s^2
M = 3225Lbs * 0.454kg/1Lb = = 1464 kg
F = M*a = 1464 * (-10.26) = -15,022 N.
Note: Your answer would have been correct if you had converted 60 mph to
m/s.
V = Vo + a*t = 0
a = -Vo/t = -26.67/2.6 = -10.26 m/s^2
M = 3225Lbs * 0.454kg/1Lb = = 1464 kg
F = M*a = 1464 * (-10.26) = -15,022 N.
Note: Your answer would have been correct if you had converted 60 mph to
m/s.
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