A car of mass 1000 kg accelerates uniformly from rest to a velocity of 54 km/hour in 5s. The average power of the engine during this period in watts is
Correct answer: B. 22500 W
- A. 2000 W
- B. 22500 W
- C. 5000 W
- D. 2250 W
Explanation
The explanation for the question is: To calculate the average power of the engine during the given period, we can use the formula: Average Power = Work / Time First, let's find the work done by the engine. The work done (W) is given by the formula: Work = Force × Distance In this case, the force is equal to the product of mass (m) and acceleration (a): Force = m × a The acceleration can be calculated using the formula: Acceleration = Change in Velocity / Time Given that the car starts from rest (initial velocity, u = 0) and reaches a final velocity (v) of 54 km/hour, we need to convert the velocity to m/s: 54 km/hour = 54 × (1000 m/3600 s) = 15 m/s Now, we can calculate the acceleration: Acceleration = (v - u) / t = (15 m/s - 0 m/s) / 5 s = 3 m/s² Substituting the values into the formula, we get: Force = m × a = 1000 kg × 3 m/s² = 3000 N Next, we need to find the distance traveled by the car. The formula for distance (s) can be calculated using the kinematic equation: s = ut + (1/2)at² Since the car starts from rest (u = 0), the equation simplifies to: s = (1/2)at² = (1/2)(3 m/s²)(5 s)² = 37.5 m Now we can calculate the work done: Work = Force × Distance = 3000 N × 37.5 m = 112,500 J Finally, we can calculate the average power: Average Power = Work / Time = 112,500 J / 5 s = 22,500 W Therefore, the average power of the engine during this period is 22,500 watts.
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About Work and Energy
Work transfers energy when a force causes displacement, while kinetic energy, gravitational potential energy and power describe motion, position and the rate of energy transfer. The work-energy theorem links net work with change in kinetic energy, and efficiency accounts for energy losses rather than treating them as energy destruction.
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