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Chapter 4: Problem 562

A particle is moving under the influence of a force given by \(\mathrm{F}=\mathrm{kx}\), where \(\mathrm{k}\) is a constant and \(\mathrm{x}\) is the distance moved. What energy (in joule) gained by the particle in moving from \(\mathrm{x}=1 \mathrm{~m}\) to \(\mathrm{x}=3 \mathrm{~m} ?\) (A) \(2 \mathrm{k}\) (B) \(3 \mathrm{k}\) (C) \(4 \mathrm{k}\) (D) \(9 \mathrm{k}\)

Short Answer

Expert verified
The energy gained by the particle in moving from \(x = 1m\) to \(x = 3m\) is (C) \(4k\).

Step by step solution

01

Recall the formula for work

The formula for work done by a force that varies with position is given by: \[W = \int_{x_1}^{x_2} F(x) dx\] In our case, \(F(x) = kx\), \(x_1 = 1m\) and \(x_2 = 3m\).
02

Calculate the work done by the force

Plug the values into the formula for work and calculate the integral: \[W = \int_{1}^{3} kx \,dx\] To integrate, use the power rule: \(\int x^n dx = \frac{1}{n+1} x^{n+1} + C\). In this case, \(n = 1\), so: \[W = k \left(\frac{1}{2} x^{2}\right)\Bigr|_{1}^{3}\]
03

Evaluate the integral

Now, evaluate the expression at the limits of integration: \[W = k\left[\frac{1}{2} (3^2) - \frac{1}{2} (1^2)\right] = k\left[\frac{9}{2} - \frac{1}{2} \right] = k\left[\frac{8}{2}\right] = 4k\] So, the work done by the force is \(4k\).
04

Calculate the energy gained by the particle

The work-energy theorem states that the work done on an object is equal to the change in its kinetic energy: \[W = \Delta KE\] In our case, we found that the work done, \(W = 4k\). Therefore, the energy gained by the particle is also equal to \(4k\). The correct answer is (C) \(4k\).

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Most popular questions from this chapter

A \(60 \mathrm{~kg}\) JATAN with \(10 \mathrm{~kg}\) load on his head climbs 25 steps of \(0.20 \mathrm{~m}\) height each. What is the work done in climbing ? \(\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)\) (A) \(5 \mathrm{~J}\) (B) \(350 \mathrm{~J}\) (C) \(100 \mathrm{~J}\) (D) \(3500 \mathrm{~J}\)

An electric motor develops \(5 \mathrm{KW}\) of power. How much time will it take to lift a water of mass \(100 \mathrm{~kg}\) to a height of $20 \mathrm{~m}\( ? \)\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)$ (A) \(4 \mathrm{sec}\) (B) \(5 \mathrm{sec}\) (C) \(8 \mathrm{sec}\) (D) \(10 \mathrm{sec}\)

The velocity of a body of mass \(400 \mathrm{gm}\) is $(-3 \mathrm{i} \wedge-4 \mathrm{j} \wedge) \mathrm{m} / \mathrm{s}$. So its kinetic energy is ...... (A) \(5 \mathrm{~J}\) (B) \(10 \mathrm{~J}\) (C) \(8 \mathrm{~J}\) (D) \(16 \mathrm{~J}\)

A single conservative force \(\mathrm{F}(\mathrm{x})\) acts on a $2.5 \mathrm{~kg}\( particle that moves along the \)\mathrm{x}$ -axis. The potential energy \(\mathrm{U}(\mathrm{x})\) is given by \(\mathrm{U}(\mathrm{x})=\left[10+(\mathrm{x}-4)^{2}\right]\) where \(\mathrm{x}\) is in meter. \(\mathrm{At} \mathrm{x}=6.0 \mathrm{~m}\) the particle has kinetic energy of \(20 \mathrm{~J}\). what is the mechanical energy of the system? (A) \(34 \mathrm{~J}\) (B) \(45 \mathrm{~J}\) (C) \(48 \mathrm{~J}\) (D) \(49 \mathrm{~J}\)

A gun fires a bullet of mass \(40 \mathrm{~g}\) with a velocity of $50 \mathrm{~m} / \mathrm{s}$. Because of this the gun is pushed back with a velocity of \(1 \mathrm{~m} / \mathrm{s}\). The mass of the gun is (A) \(1.5 \mathrm{~kg}\) (B) \(3 \mathrm{~kg}\) (C) \(2 \mathrm{~kg}\) (D) \(2.5 \mathrm{~kg}\)
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