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Chapter 4: Problem 491

A body initially at rest undergoes one dimensional motion with constant acceleration. The power delivered to it at time \(\mathrm{t}\) is proportional to..... (A) \(\mathrm{t}^{1 / 2}\) (B) \(t\) (C) \(\mathrm{t}^{3 / 2}\) (D) \(\mathrm{t}^{2}\)

Short Answer

Expert verified
The power delivered to the body at time \(t\) is proportional to \(t\).

Step by step solution

01

Identify the known values

Since the body is initially at rest, it is given that Initial velocity, \(v_0 = 0\). Also, it undergoes constant acceleration, so there exists a constant acceleration \(a\). We need to find the power delivered to the body at time \(t\).
02

Find the formula for work done (w)

The work-energy theorem states that the work done (W) on the body equals the change in its kinetic energy. \[W = \frac{1}{2}m(v^2 - v_0^2)\] Since \(v_0 = 0\), we get: \[W = \frac{1}{2}mv^2\]
03

Express velocity (v) in terms of acceleration (a) and time (t)

From the equation of motion with constant acceleration, \(v = v_0 + at\) Since initial velocity \(v_0 = 0\), we have: \(v = at\)
04

Substitute the expression for velocity in the work done formula

Replace \(v\) with \(at\) in the work done formula: \[W = \frac{1}{2}m(at)^2 = \frac{1}{2}ma^2t^2\]
05

Calculate the power (P) formula

The power (P) is the rate at which work is done, so we need to find the derivative of the work with respect to time (t). \[P = \frac{dW}{dt}\] Differentiating the work formula with respect to time: \[P = \frac{d}{dt}\left(\frac{1}{2}ma^2t^2\right) = ma^2t\]
06

Find the proportion of power to time

Consider the power formula \(P = ma^2t\). We need to determine which of the given options best represents the proportionality between power and time. Comparing the formula with each option, we get that \(P \propto t^{1}\) Therefore, the correct option is: (B) \(t\)

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Most popular questions from this chapter

A sphere of mass \(\mathrm{m}\) moving the velocity \(\mathrm{v}\) enters a hanging bag of sand and stops. If the mass of the bag is \(\mathrm{M}\) and it is raised by height \(\mathrm{h}\), then the velocity of the sphere was (A) \([\\{\mathrm{m}+\mathrm{M}\\} / \mathrm{m}] \sqrt{(2 \mathrm{gh})}\) (B) \((\mathrm{M} / \mathrm{m}) \sqrt{(} 2 \mathrm{gh})\) (C) \([\mathrm{m} /\\{\mathrm{M}+\mathrm{m}\\}] \sqrt{(2 \mathrm{gh})}\) (D) \((\mathrm{m} / \mathrm{M}) \sqrt{(2 \mathrm{gh})}\)

A spring is compressed by \(1 \mathrm{~cm}\) by a force of \(4 \mathrm{~N}\). Find the potential energy of the spring when it is compressed by \(10 \mathrm{~cm}\) (A) \(2 \mathrm{~J}\) (B) \(0.2 \mathrm{~J}\) (C) \(20 \mathrm{~J}\) (D) \(200 \mathrm{~J}\)

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) \(3500 \mathrm{~J}\)

A cord is used to lower vertically a block of mass \(\mathrm{M}\) by a distance \(\mathrm{d}\) with constant downward acceleration \((9 / 2)\). Work done by the cord on the block is (A) \(-\mathrm{Mgd} / 2\) (B) \(\mathrm{Mgd} / 4\) (C) \(-3 \mathrm{Mgd} / 4\) (D) \(\mathrm{Mgd}\)

From an automatic gun a man fires 240 bullet per minute with a speed of $360 \mathrm{~km} / \mathrm{h}\(. If each weighs \)20 \mathrm{~g}$, the power of the gun is (A) \(400 \mathrm{~W}\) (B) \(300 \mathrm{~W}\) (C) \(150 \mathrm{~W}\) (D) \(600 \mathrm{~W}\)
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