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Chapter 4: Problem 447

A force \(F=A y^{2}+B y+C\) acts on a body in the \(y\) -direction. The work done by this force during a displacement from \(\mathrm{y}=-\mathrm{a}\) to \(\mathrm{y}=\mathrm{a}\) is (A) \(\left[\left\\{2 \mathrm{Aa}^{3}\right\\} / 3\right]\) (B) \(\left[\left\\{2 \mathrm{Aa}^{3}\right\\} / 3\right]+2 \mathrm{ca}\) (C) $\left[\left\\{2 \mathrm{Aa}^{3}\right\\} / 3\right]+\left[\left\\{\mathrm{Ba}^{2}\right\\} / 2\right]+\mathrm{ca}$ (D) None of these.

Short Answer

Expert verified
The short answer for the work done is: (B) \(\left[\left\\{2 \mathrm{Aa}^{3}\right\\} / 3\right]+2 \mathrm{ca}\)

Step by step solution

01

Identify the given information

The force acting on the body in the y-direction is given by F = Ay² + By + C, and we need to find the work done during the displacement from y = -a to y = a.
02

Write the formula for work done

The work done (W) by a force acting in the y-direction can be calculated by integrating the force F with respect to y over the given interval [-a, a]: \[W = \int_{-a}^{a} F(y) dy\]
03

Substitute the expression for F

Replace F in the integral expression with Ay² + By + C: \[W = \int_{-a}^a (Ay^2 + By + C) dy\]
04

Integrate the expression

Now, integrate each term with respect to y: \[W = \int_{-a}^a Ay^2 dy + \int_{-a}^a By dy + \int_{-a}^a C dy\] \[W = A\int_{-a}^a y^2 dy + B\int_{-a}^a y dy + C\int_{-a}^a dy\] For each integral: \[\int y^2 dy = \frac{1}{3} y^3 + D_1\] \[\int y dy = \frac{1}{2} y^2 + D_2\] \[\int dy = y + D_3\] Integration constants D1, D2, and D3 are not necessary, so the results are: \[W = A\left[\frac{1}{3} y^3\right]_{-a}^a + B\left[\frac{1}{2} y^2\right]_{-a}^a + C[y]_{-a}^a\]
05

Evaluate the integral at the endpoints

Now, evaluate the expression at the endpoints y = -a and y = a: \[W = A\left(\frac{1}{3}(a^3) - \frac{1}{3}(-a)^3\right) + B\left(\frac{1}{2}(a^2) - \frac{1}{2}(-a)^2\right) + C(a - (-a))\]
06

Simplify the expression

Simplify the expression for the work done: \[W = A\left(\frac{2a^3}{3}\right) + B\left(\frac{a^2}{2} - \frac{a^2}{2}\right) + C(2a)\] \[W = \frac{2Aa^3}{3} + 0 + 2Ca\]
07

Choose the correct answer

The simplified expression for the work done is: \[W = \frac{2Aa^3}{3} + 2Ca\] Comparing this with the given choices, we find that the correct answer is (B) \(\left[\left\\{2 \mathrm{Aa}^{3}\right\\} / 3\right]+2 \mathrm{ca}\).

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Most popular questions from this chapter

Assertion and Reason are given in following questions. Each question have four option. One of them is correct it. (1) If both assertion and reason and the reason is the correct explanation of the Assertion. (2) If both assertion and reason are true but reason is not the correct explanation of the assertion. (3) If the assertion is true but reason is false. (4) If the assertion and reason both are false. Assertion: stopping distance \(=[\\{\) Kinetic energy \(\\} /\\{\) Stopping force \(\\}]\) Reason: Work done in stopping a body is equal to K.E. of the body. (A) 1 (B) 2 (C) 3 (D) 4

How much is the work done in pulling up a block of wood weighing $2 \mathrm{KN}\( for a length of \)10 \mathrm{~m}$ on a smooth plane inclined at an angle of \(30^{\circ}\) with the horizontal? (A) \(1.732 \mathrm{KJ}\) (B) \(17.32 \mathrm{KJ}\) (C) \(10 \mathrm{KJ}\) (D) \(100 \mathrm{KJ}\)

The force constant of a wire is \(\mathrm{K}\) and that of the another wire is \(3 \mathrm{k}\) when both the wires are stretched through same distance, if work done are \(\mathrm{W}_{1}\) and \(\mathrm{W}_{2}\), then... (A) \(\mathrm{w}_{2}=3 \mathrm{w}_{1}^{2}\) (B) \(\mathrm{W}_{2}=0.33 \mathrm{~W}_{1}\) (C) \(\mathrm{W}_{2}=\mathrm{W}_{1}\) (D) \(\mathrm{W}_{2}=3 \mathrm{~W}_{1}\)

The mass of a car is \(1000 \mathrm{~kg}\). How much work is required to be done on it to make it move with a speed of \(36 \mathrm{~km} / \mathrm{h}\) ? (A) \(2.5 \times 10^{4} \mathrm{~J}\) (B) \(5 \times 10^{3} \mathrm{~J}\) (C) \(500 \mathrm{~J}\) (D) \(5 \times 10^{4} \mathrm{~J}\)

A uniform chain of length \(\mathrm{L}\) and mass \(\mathrm{M}\) is lying on a smooth table and \((1 / 4)^{\text {th }}\) of its length is hanging vertically down over the edge of the table. If \(g\) is acceleration due to gravity, the work required to pull the hanging part on to the table is (A) MgL (B) \(\mathrm{MgL} / 9\) (C) \(\mathrm{MgL} / 18\) (D) \(\mathrm{MgL} / 32\)
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