A 1.54×10
3
kg car accelerates uniformly from
rest to 12.7 m/s in 2.97 s.
What is the work done on the car in this
time interval?
Answer in units of J
002 (part 2 of 2) 10.0 points
What is the power delivered by the engine in
this time interval?
Answer in units of W
To find the work done on the car, we can use the work-energy principle. The work done on an object is equal to the change in its kinetic energy.
The formula for work is given by:
Work = ΔKE = KE_final - KE_initial
To calculate the initial and final kinetic energy of the car, we need to use the mass of the car and its final velocity.
Given:
Mass of the car (m) = 1.54×10^3 kg
Final velocity (v) = 12.7 m/s
The initial velocity is zero since the car starts from rest. Thus, the initial kinetic energy (KE_initial) is zero.
To calculate the final kinetic energy (KE_final), we use the formula:
KE = 0.5 * mass * velocity^2
KE_final = 0.5 * m * v^2
Now we can substitute the given values into the formula:
KE_final = 0.5 * (1.54×10^3 kg) * (12.7 m/s)^2
Calculate KE_final:
KE_final = 0.5 * 1.54×10^3 kg * (161.29 m^2/s^2)
Next, we find the change in kinetic energy (ΔKE):
ΔKE = KE_final - KE_initial
Since the initial kinetic energy is zero, we have:
ΔKE = KE_final - 0
Finally, the work done on the car is equal to the change in kinetic energy:
Work = ΔKE = KE_final
Therefore, the work done on the car is equal to the final kinetic energy:
Work = 0.5 * 1.54×10^3 kg * (161.29 m^2/s^2)
Calculate the work to find your answer in units of Joules (J).
To calculate the power delivered by the engine, we can use the formula:
Power = Work / Time
Given:
Time (t) = 2.97 s
Substitute the work value we calculated and the time value into the formula:
Power = Work / Time
Calculate the power to find your answer in units of Watts (W).