A 800 kg car is travelling forwards at + 25 m s-1 along a level road. The brake

Question

A 800 kg car is travelling forwards at + 25 m s-1 along a level road. The brake is suddenly applied, and the car skids to a stop leaving skid marks 75 m long. The acceleration (assumed constant) during skidding will be closest to 0.2 m s 8.3 m s-2 -4.2 m s-2 -8.3 m s -2 The shape of the DISPLACEMENT (s) versus time (t) graph for this process will be closest to During braking, which of the following statements will be true? The net force on the car is unbalanced. The frictional force of the car on the ground is directed forwards. There are no forces on the car which are directed forwards. All of these are true. Which of the following pairs of forces represents an action-reaction pair? The weight of the car and the force of gravity exerted by the car on the earth. The weight of the car and the normal force of the car on the road. The weight of the car and the normal force of the road on the car. All of these are action-reaction pairs. The magnitude of the work done by friction in bringing the car to a stop will be 6.7 kJ 588 kJ 250 kJ 3.3 kJ unable to be calculated without the coefficient of friction.

Explanation / Answer

1.) option C

v2 - u2 = 2as here u = 0 v = 25m/s s = 75m

a = (0-625) / 150 = -4.16m/s2  approx -4.2m/s2

2.) option B

s = ut + 0.5 at2

here s = 25t - 2.1t2

this is a downward facing parabola.. at t=0 s = 0 so it starts from origin

3.) option D

during breaking velocity is lost implies there is decelaration so there must be an unbalanced force F= m (-a)

friction on the car acts backwards so the friction by the car on the ground wil act forward

forces acting on the car are only friction and normal force.. neither of them act in forward direction

4.) option C, A

option A is correct as both forces mentioned act in opposite directions and equal so action and reaction pair

option B is wrong as normal force of the car on road is nothing but the weight of the car.. they are same

option C is correct.. as weight of car acts on road, road exerts a counter force ie normal force on the car

5.) option C

work done = change in Kinectic energy

W = 0.5mv2 - 0.5mu2

W = 0.5 * 800 * 252 - 0.5 *800 * 0 = 250KJ

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