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Chapter 4: Problem 494

A rope-way trolley of mass \(1200 \mathrm{~kg}\) uniformly from rest to a velocity of \(72 \mathrm{~km} / \mathrm{h}\) in \(6 \mathrm{~s}\). What is the average power of the engine during this period in watt ? (Neglect friction) (A) \(400 \mathrm{~W}\) (B) \(40,000 \mathrm{~W}\) (C) \(24000 \mathrm{~W}\) (D) \(4000 \mathrm{~W}\)

Short Answer

Expert verified
The average power of the engine during this period is 40,000 W. The correct answer is (B) \(40,000 \mathrm{~W}\).

Step by step solution

01

Convert the given velocity from km/h to m/s

To convert the velocity from km/h to m/s, we use the conversion factor 1000/3600 (1 km = 1000 m, 1 hour = 3600 seconds): \[v = 72 \frac{\text{km}}{\text{h}} \cdot \frac{1000 \text{ m}}{1 \text{ km}} \cdot \frac{1 \text{ h}}{3600 \text{ s}} = 20 \frac{\text{m}}{\text{s}}\]
02

Calculate the final kinetic energy (K.E.) of the trolley

Now, we need to calculate the final kinetic energy of the trolley. The formula for kinetic energy is: \[ K.E. = \frac{1}{2}mv^2\] Using the given mass (1200 kg) and calculated velocity (20 m/s), we get: \[ K.E. = \frac{1}{2}\times 1200 \times (20)^2 = 240,000 \mathrm{~J}\]
03

Calculate the work done (W) on the trolley

The trolley started from rest so its initial kinetic energy was zero. The work done on the trolley is the difference between its final and initial kinetic energies. In this case, the initial K.E. is zero so the difference is the same as the final K.E.: \[W = K.E. - 0 = 240,000 \mathrm{~J}\]
04

Calculate the average power (P) of the engine

The average power (P) of the engine can be calculated using the formula: \[P = \frac{W}{t}\] where t is the time taken. We know t=6s and W=240,000J. \[P = \frac{240,000 \mathrm{~J}}{6 \mathrm{~s}} = 40,000 \mathrm{~W}\] The average power of the engine during this period is 40,000 W. The correct answer is (B) \(40,000 \mathrm{~W}\).

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Most popular questions from this chapter

An ice-cream has a marked value of \(700 \mathrm{kcal}\). How many kilo-watt- hour of energy will it deliver to the body as it is digested $(\mathrm{J}=4.2 \mathrm{~J} / \mathrm{cal})$ (A) \(0.81 \mathrm{kwh}\) (B) \(0.90 \mathrm{kwh}\) (C) \(1.11 \mathrm{kwh}\) (D) \(0.71 \mathrm{kwh}\)

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