Answer:
35.6 kW
Explanation:
Given data
mass (m): 1400 kg
speed (v): 60.0 mi/h. In SI units,
[tex]\frac{60.0mi}{h}.\frac{1609.34m}{1mi}.\frac{1h}{3600s} =26.8m/s[/tex]
time (t): 14.0 s
The acceleration (a) of the car is:
[tex]a=\frac{\Delta v}{t} =\frac{26.8m/s}{14.0s} =1.91m/s^{2}[/tex]
The force (F) exerted is:
F = m × a = 1400 kg × 1.91 m/s² = 2.67 × 10³ N
The distance (d) travelled is:
d = 1/2 × a × t² = 1/2 × 1.91 m/s² × (14.0 s)² = 187 m
The work (w) done is:
w = F × d = 2.67 × 10³ N × 187 m = 4.99 × 10⁵ J
The power (P) of the car engine is:
P = w / t = 4.99 × 10⁵ J / 14.0 s = 3.56 × 10⁴ W = 35.6 kW
The power the car engine that contributes to speeding is 48.16 hp.
Given the data in the question;
Power a car engine; [tex]P =\ ?[/tex]
Power is the amount energy transferred per unit time.
[tex]Power = \frac{Work\ done }{time}[/tex]
From the work-energy theorem:
Work done on an object by a net force is equals to the change in kinetic energy of that object.
[tex]W = KE_f - KE_i[/tex]
Hence, [tex]Power = \frac{KE}{time} = \frac{\frac{1}{2}mv^2 }{t}[/tex]
We substitute in our given values
[tex]P = \frac{\frac{1}{2}*1400kg*(26.8m)^2 }{14.0s} \\\\P = \frac{502768kg.m^2}{14.0s} \\\\P = 35912 W\\\\P = 48.16 hp[/tex]
Therefore, the power the car engine that contributes to speeding is 48.16 hp.
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