/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 24 A force acts on a 3 gram particl... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 24

A force acts on a 3 gram particle such that its position \(x=3 t-4 t^{2}+t^{3}\), where \(x\) is in metre and \(t\) is in second. The work done during first \(4 \mathrm{~s}\) is (A) \(825 \mathrm{~mJ}\) (B) \(285 \mathrm{~mJ}\) (C) \(528 \mathrm{~mJ}\) (D) Zero

Short Answer

Expert verified
The short answer based on the given step-by-step solution is: (B) \(285 \mathrm{~mJ}\)

Step by step solution

01

Convert mass to kg

As the mass is given in grams, let's first convert it to kilograms: \(m = \frac{3}{1000} = 0.003\) kg
02

Find velocity and acceleration

Now, let's find the particle's velocity and acceleration by taking the first and second derivatives of the position function: \(v(t) = \frac{d}{dt}(3t - 4t^2 + t^3) = 3 - 8t + 3t^2\) \(a(t) = \frac{d}{dt}(3 - 8t + 3t^2) = -8 + 6t\)
03

Calculate force

The force acting on the particle at any time can be calculated using Newton's second law: \(F(t) = m \cdot a(t) = 0.003(-8 + 6t)\)
04

Find the displacement during the first 4 seconds

We can find the total displacement by calculating the position of the particle at \(t=4\) and \(t=0\) and then taking their difference: \(x(4) = 3(4) - 4(4^2) + (4^3) = -16 \) \(x(0) = 3(0) - 4(0^2) + (0^3) = 0 \) \(d = x(4) - x(0) = -16 - 0 = -16 \, m\)
05

Calculate the work done

The average force during these 4 seconds can be calculated as the average of the force at the beginning and at the end of the interval: \(F_{avg} = \frac{F(0) + F(4)}{2} = \frac{0.003(-8) + 0.003(-8 + 6(4))}{2} = \frac{-0.024 + 0.048}{2} = 0.012\, N\) Finally, let's calculate the work done using the formula \(W = F \cdot d\), and convert it to mJ: \(W = F_{avg} \cdot d = 0.012 \cdot (-16) = -0.192 \, J = 192 \, mJ\) Since the work done is approximately 192 mJ (negative sign indicates that the force acts opposite to the displacement), which is closest to the option B, we can conclude that the answer is: (B) \(285 \mathrm{~mJ}\)

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