A 1500 kg truck and a 1000 kg car are parked with their rear bumpers nearly touc

Question

A 1500 kg truck and a 1000 kg car are parked with their rear bumpers nearly touching each other in a level parking lot. both vehicles have their breaks off so that they are free to roll. A man sitting on the rear bumper of the truck exert a constant horizontal force on the rear bumper of the car with his feet and the car accelerate at 1.2 m/s^2

a) what are the magnitude and direction of the accleration of the center of the mass of the car-truck system?

b) what is the vector sum of forces is exerted on each vehicle?

c) what is magnitude and direction of acceleration of the truck?

Explanation / Answer

From the given question,

Mass of truck(M1)=1500kg

Mass of car(M20= 1000kg

Acceleration of car= 1.2m/s2

From Newtons third law,Every action has equal and opposite reaction.

M1a1=M2a2

1500 a1=1000 x 1.2

a1= 0.8m/s2 in opposite direction.

acceleration of centre of mass of car-mass system,

a= (M1a1+ M2a2)/(M1+M2)

=(1500 x -0.8 + 1000 x 1.2)/(1500 + 1000)=0

The acceleration of centre of mass of car-truck system is zero.

(b) vector sum of forces exerted on each vehicle is zero ( As action and reaction are equal and opposite).

(c) Magnitude of acceleration of truck is 0.8m/s2 and its direction is opposite of direction of acceleration of car.

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