The 400 kg mine car is hoisted up the incline using the cable and motor m. for a short time, the force in the cable is f= (3200) n, where t is in seconds. if the car has an initial velocity v1= 2m/s at s 0 and t= 0, determine the distance it moves the plane when (a) t 1 s and (b) f-5 s.

(a) 110 m/s

(b) 42 m/s

Explanation:

mass, m = 400 kg, F = 3200 N, V1 = 2 m/s,

acceleration, a = Force / mass = 3200 / 400 = 8 m/s^2

(a) Use first equation of motion

v = V1 + a t

v = 2 + 8 x 1 = 10 m/s

(b) Again using first equation of motion

v = V1 + a t

v = 2 + 8 x 5 = 42 m/s

Thus, the velocity of plane after 1 second is 10 m/s and after 5 second the velocity is 42 m/s.

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