A car speeding along a level road at 25m/s. the brakes abruptly bring the 500kg car to a speed of 3m/s over distance opf 5m. determine the force allied by brakes and the work done by the car on the brakes.
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F = ma where m is the mass of the body and a the acceleration of the body.
Given the distance, the initial and the final velocity, we can get the acceleration.
Calculating time:
Time = Distance ÷ speed
Time =5/3=1.6667seconds.
Acceleration = Δ in velocity /Δ in time.
Δ in velocity = Final velocity - initial velocity
= (3 - 25)=-22m/s
Acceleration =22/1.6667=-13.1997m/s²
This means the car is decelerating.
F =ma
We take the absolute value since force is positive.
F =500 × 13.1997 =6599.85N
Work done = -Force × distance since the work is done in the opposite direction of the motion of the car.
-6599.85 × 5 = - 32999.25 Joules.
Given the distance, the initial and the final velocity, we can get the acceleration.
Calculating time:
Time = Distance ÷ speed
Time =5/3=1.6667seconds.
Acceleration = Δ in velocity /Δ in time.
Δ in velocity = Final velocity - initial velocity
= (3 - 25)=-22m/s
Acceleration =22/1.6667=-13.1997m/s²
This means the car is decelerating.
F =ma
We take the absolute value since force is positive.
F =500 × 13.1997 =6599.85N
Work done = -Force × distance since the work is done in the opposite direction of the motion of the car.
-6599.85 × 5 = - 32999.25 Joules.
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