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Chapter 4: Problem 428

A body of mass \(6 \mathrm{~kg}\) is under a force, which causes a displacement in it given by $\mathrm{S}=\left[\left(2 \mathrm{t}^{3}\right) / 3\right](\mathrm{in} \mathrm{m}) .$ Find the work done by the force in first one seconds. (A) \(2 \mathrm{~J}\) (B) \(3.8 \mathrm{~J}\) (C) \(5.2 \mathrm{~J}\) (D) \(24 \mathrm{~J}\)

Short Answer

Expert verified
The work done by the force in the first one second is 12 J. The correct answer is (D).

Step by step solution

01

Find the velocity function

To find the velocity function, we need to differentiate the given displacement function with respect to time. Given, S(t) = (2t³) / 3 Differentiate S(t) with respect to t: v(t) = dS/dt = d(2t³/3)/dt = 2t²
02

Find the acceleration function

To find the acceleration function, we need to differentiate the velocity function with respect to time. v(t) = 2t² Differentiate v(t) with respect to t: a(t) = dv/dt = d(2t²)/dt = 4t
03

Find the force function

To find the force function, we should substitute the value of mass (m = 6 kg) and the acceleration function in F = m*a. a(t) = 4t F(t) = m*a(t) = 6*(4t) = 24t
04

Find the work done during the first second

To find the work done during the first second, we need to integrate the force function over the time interval from 0 to 1 second. W = ∫(F(t)dt) from 0 to 1 W = ∫(24t*dt) from 0 to 1 Now, integrate the function: W = (12t²) | from 0 to 1 = 12(1)² - 12(0)² = 12 J So, the work done by the force in the first one second is 12 J. The correct answer is (D) 12 J.

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Most popular questions from this chapter

A body of mass \(1 \mathrm{~kg}\) is thrown upwards with a velocity $20 \mathrm{~m} / \mathrm{s}\(. It momentarily comes to rest after a height \)18 \mathrm{~m}\(. How much energy is lost due to air friction. \)(\mathrm{g}=10 \mathrm{~m} / \mathrm{s} 2)$ (A) \(20 \mathrm{~J}\) (B) \(30 \mathrm{~J}\) (C) \(40 \mathrm{~J}\) (D) \(10 \mathrm{~J}\)

A particle is acted upon by a force \(\mathrm{F}\) which varies with position \(\mathrm{x}\) as shown in figure. If the particle at \(\mathrm{x}=0\) has kinetic energy of \(20 \mathrm{~J}\). Then the calculate the kinetic energy of the particle at \(\mathrm{x}=16 \mathrm{~cm}\). (A) \(45 \mathrm{~J}\) (B) \(30 \mathrm{~J}\) (C) \(70 \mathrm{~J}\) (D) \(135 \mathrm{~J}\)

Natural length of a spring is \(60 \mathrm{~cm}\), and its spring constant is \(2000 \mathrm{~N} / \mathrm{m}\). A mass of \(20 \mathrm{~kg}\) is hung from it. The extension produced in the spring is..... $\left(\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^{2}\right)$ (A) \(4.9 \mathrm{~cm}\) (B) \(0.49 \mathrm{~cm}\) (C) \(9.8 \mathrm{~cm}\) (D) \(0.98 \mathrm{~cm}\)

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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) \(14 \mathrm{~m}\)
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