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Chapter 4: Problem 427

The mass of a car is \(1000 \mathrm{~kg}\). How much work is required to be done on it to make it move with a speed of \(36 \mathrm{~km} / \mathrm{h}\) ? (A) \(2.5 \times 10^{4} \mathrm{~J}\) (B) \(5 \times 10^{3} \mathrm{~J}\) (C) \(500 \mathrm{~J}\) (D) \(5 \times 10^{4} \mathrm{~J}\)

Short Answer

Expert verified
The work required to make the car move with a speed of \(36 \mathrm{~km} / \mathrm{h}\) is (D) \(5 \times 10^{4} \mathrm{~J}\).

Step by step solution

01

Convert the speed to m/s

First, we convert the speed from km/h to m/s by multiplying it by 1000 to convert kilometers to meters, then dividing by 3600 to convert hours to seconds. So, we have: \[ v = \frac{36 \text{ km/h} \times 1000 \text{m/km}}{3600 \text{s/h}} = 10 \text{ m/s} \]
02

Calculate the kinetic energy

Now that we have the speed in m/s, we can plug the mass (m = 1000 kg) and speed (v = 10 m/s) into the kinetic energy formula: \[ KE = \frac{1}{2}mv^2 = \frac{1}{2} \times 1000\text{ kg} \times (10\text{ m/s})^2 \]
03

Compute the work done

Since work done is equal to the transfer of energy, and in this case, it is the transfer of kinetic energy, we can find the work done by solving the equation from step 2: \[ W = KE = \frac{1}{2} \times 1000\text{ kg} \times (10\text{ m/s})^2 = \frac{1}{2} \times 1000 \times 100 = 50,\!000\text{ J} \] From the given options, the work required to make the car move with a speed of 36 km/h is (D) \(5 \times 10^{4} \mathrm{~J}\).

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Most popular questions from this chapter

A metal ball of mass \(2 \mathrm{~kg}\) moving with a velocity of $36 \mathrm{~km} / \mathrm{h}$ has a head on collision with a stationary ball of mass \(3 \mathrm{~kg}\). If after the collision, the two balls move together, the loss in kinetic energy due to collision is (A) \(40 \mathrm{~J}\) (B) \(60 \mathrm{~J}\) (C) \(100 \mathrm{~J}\) (D) \(140 \mathrm{~J}\)

A bomb of \(12 \mathrm{~kg}\) divides in two parts whose ratio of masses is $1: 4 .\( If kinetic energy of smaller part is \)288 \mathrm{~J}$, then momentum of bigger part in \(\mathrm{kgm} / \mathrm{sec}\) will be (A) 38 (B) 72 (C) 108 (D) Data is incomplete

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