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Chapter 4: Problem 450

With what velocity should a student of mass \(40 \mathrm{~kg}\) run so that his kinetic energy becomes \(160 \mathrm{~J}\) ? (A) \(4 \mathrm{~m} / \mathrm{s}\) (B) \(\sqrt{8} \mathrm{~m} / \mathrm{s}\) (C) \(16 \mathrm{~m} / \mathrm{s}\) (D) \(8 \mathrm{~m} / \mathrm{s}\)

Short Answer

Expert verified
The student should run with a velocity of \(\sqrt{8} \mathrm{~m} / \mathrm{s}\).

Step by step solution

01

Write down the kinetic energy equation

First, let's write down the equation relating kinetic energy (KE), mass (m), and velocity (v): \[KE = \frac{1}{2}mv^2\] Since we are given the mass (40 kg) and the desired kinetic energy (160 J), we can plug these values into the equation and solve for v.
02

Plug in the given values

Now, let's plug in the given values for mass (m) and kinetic energy (KE) into the equation: \[160 J = \frac{1}{2}(40 kg)v^2\]
03

Solve for velocity (v)

To find the velocity (v), we can follow these steps: 1. Divide both sides by 40 kg: \[\frac{160 J}{40 kg} = \left(\frac{1}{2}v^2\right)\] 2. Multiply both sides by 2: \[\frac{320 J}{40 kg} = v^2\] 3. Simplify: \[8 = v^2\] 4. Take the square root of both sides: \[\sqrt{8} = v\] So, the velocity with which the student should run is \(\sqrt{8} \mathrm{~m} / \mathrm{s}\). The correct answer is (B).

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Most popular questions from this chapter

A ball is allowed to fall from a height \(20 \mathrm{~m}\). If there is \(30 \%\) loss of energy due to impact, then after one impact ball will go up to (A) \(18 \mathrm{~m}\) (B) \(16 \mathrm{~m}\) (C) \(12 \mathrm{~m}\) (D) \(14 \mathrm{~m}\)

The decreases in the potential energy of a ball of mass \(25 \mathrm{~kg}\) which falls from a height of \(40 \mathrm{~cm}\) is (A) \(968 \mathrm{~J}\) (B) \(100 \mathrm{~J}\) (C) \(1980 \mathrm{~J}\) (D) \(200 \mathrm{~J}\)

The bob of simple pendulum (mass \(\mathrm{m}\) and length 1 ) dropped from a horizontal position strike a block of the same mass elastically placed on a horizontal frictionless table. The K.E. of the block will be (A) \(2 \mathrm{mg} 1\) (B) \(\mathrm{mg} 1 / 2\) (C) \(\mathrm{mg} 1\) (D) zero

When \(2 \mathrm{~kg}\) mass hangs to a spring of length \(50 \mathrm{~cm}\), the spring stretches by \(2 \mathrm{~cm}\). The mass is pulled down until the length of the spring becomes \(60 \mathrm{~cm}\). What is the amount of elastic energy stored in the spring in this condition, if $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$ (A) \(10 \mathrm{~J}\) (B) \(2 \mathrm{~J}\) (C) \(2.5 \mathrm{~J}\) (D) \(5 \mathrm{~J}\)

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