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Sagot :
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
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
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
Learn more about energy:
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