Question

A 1,000 kg car traveling 15.0 m/s brakes and comes to a stop after traveling 20.0 m.


a. What is the car’s initial kinetic energy?
b. What is the car’s final kinetic energy?
c. How much work does it take to stop the car?
d. How much constant force is applied in bringing the car to a stop?

Answer 1

a. initial kinetic
Ek=1/2mv ²
Ek=1/2(1000)(15²)
Ek=112,500J

b. final kinetic energy=0J
as it comes to a stop so v=0

c. Work = ∆Energy
Or
Work=Fs

∆E/W= initial energy - final energy
∆E/W=112,500-0
∆E/W=112,500 J

d.W=Fs

112,500=F(20)
112,500/20 =F
F=5625 N

Answer 2

A. The car’s initial kinetic energy is 112500 J

B. The car’s final kinetic energy is 0 J

C. The amount of work done to stop the car is –112500 J

D. The force applied in bringing the car to a stop is –5625 N

A. How to determine the initial kinetic energy

• Mass (m) = 1000 Kg
• Initial velocity (u) = 15 m/s
• Initial kinetic energy (KEᵢ) =?

KEᵢ = ½mu²

KEᵢ = ½ × 1000 × 15²

KEᵢ = 112500 J

B. How to determine the final kinetic energy

• Mass (m) = 1000 Kg
• Final velocity (v) = 0 m/s
• Final kinetic energy (KEբ) =?

KEբ = ½mv²

KEբ = ½ × 1000 × 0²

KEբ = 0 J

C. How to determine the workdone

• Initial kinetic energy (KEᵢ) = 112500 J
• Final kinetic energy (KEբ) = 0 J

• Workdone (Wd) =?

Wd = KEբ – KEᵢ

Wd = 0 – 112500

Wd = –112500 J

D. How to determine the force

We'll begin by calculating the acceleration. This can be obtained as follow:

• Initial velocity (u) = 15 m/s
• Final velocity (v) = 0 m/s
• Distance (s) = 20 m
• Acceleration (a) =?

v² = u² + 2as

0² = 15² + (2 × a × 20)

0 = 225 + 40a

Collect like terms

0 – 225 = 40a

–225 = 40a

Divide both side by 40

a = –225 / 40

a = –5.625 m/s²

Finally, we shall determine the force. This is illustrated below:

• Mass (m) = 1000 Kg
• Acceleration (a) = –5.625 m/s²

• Force (F) =?

F = ma

F = 1000 × –5.625

F = –5625 N

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