Determine the smallest radius of an unbalanced curve around which a 1600 kg car

Question

Determine the smallest radius of an unbalanced curve around which a 1600 kg car can safely travel at a constant speed of 24 m/s. The coefficient of static friction between the tires and the road is 0.60. A 68 kg Jogger runs up a long flight of stairs in 8.0 s. If the height of the stairs is 9.0 m, determine the Jogger's power output in horsepowers. How much work is required to accelerate a 1600 kg car from 20 m/s to 25 m/s? A 20 metric ton boxcar rolling on a straight track at 12 m/s strikes an identical boxcar at rest. If the cars remain locked together as a result of the collision, what is the common speed of the cars afterward? A 1600 kg car is moving at a speed of 24 m/s when the brakes are fully applied, causing all four wheels to skid. Use the work-energy-principle to determine the distance traveled by the car as it comes to a stop. The coefficient of kinetic friction between the tires and the road surface is 0.55. Use the impulse-momentum principle to find the time required for the car in problem 5 to come a stop.

Explanation / Answer

(1) mv^2 / R = u*mg

=> R = v^2 / u*g = 24^2 / (0.6*9.81) = 97.9 meter = Minimum radius.

(2) Work done in 8.0 s = mgh = 68*9.81*9.0 = 6003.72 J

Power = Work done / time = 6003.72 / 8 = 750.46 W

So power output in horse-power = 1.0 HP.

(3) Apply the formula -

v^2 = u^2 + 2*a*s

=> 2*a*s = v^2 - u^2

Work required = F*s = m*a*s = m*(v^2 - u^2) / 2 = 1600*(25^2 - 20^2) / 2 = 180000 J

(4) Common speed after collision = v / 2 = 12/2 = 6 m/s.

(5) Retardation = u*g = 0.55*9.81 = 5.4 m/s^2

So, distance travelled by the car before coming to stop, s = v^2 / 2a = 24^2 / (2*5.4) = 53.4 meter.

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