Answered

Calculate the power a car engine contributes to speeding the 1400 kg car up to a highway speed of 60.0 mi/h in 14.0 s.

Verified Answer :

dsdrajlin

Answer:

35.6 kW

Explanation:

Given data

mass (m): 1400 kg

speed (v): 60.0 mi/h. In SI units,

\(\frac{60.0mi}{h}.\frac{1609.34m}{1mi}.\frac{1h}{3600s} =26.8m/s\)

time (t): 14.0 s

The acceleration (a) of the car is:

\(a=\frac{\Delta v}{t} =\frac{26.8m/s}{14.0s} =1.91m/s^{2}\)

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

nuhulawal20

The power the car engine that contributes to speeding is 48.16 hp.

Given the data in the question;

  • Mass of car; \(m = 1400kg\)
  • Velocity; \(v = 60.0 mi/h = 26.8m/s\)
  • Time taken; \(t = 14.0s\)

Power a car engine; \(P =\ ?\)

Power is the amount energy transferred per unit time.

\(Power = \frac{Work\ done }{time}\)

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.  

\(W = KE_f - KE_i\)

Hence, \(Power = \frac{KE}{time} = \frac{\frac{1}{2}mv^2 }{t}\)

We substitute in our given values

\(P = \frac{\frac{1}{2}*1400kg*(26.8m)^2 }{14.0s} \\\\P = \frac{502768kg.m^2}{14.0s} \\\\P = 35912 W\\\\P = 48.16 hp\)

Therefore, the power the car engine that contributes to speeding is 48.16 hp.

Learn more; https://brainly.com/question/505855

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