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Chapter 4: Problem 424

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) \(\mathrm{W}_{2}=3 \mathrm{~W}_{1}\)

Short Answer

Expert verified
\(W_2 = 3\cdot W_1\)

Step by step solution

01

Write down the work done formula for each wire

For the wire with force constant K and the work done W1, the formula is: \(W_1 = \frac{1}{2}Kx^2\) For the wire with force constant 3K and the work done W2, the formula is: \(W_2 = \frac{1}{2}(3K)x^2\)
02

Divide both equations to find the relationship between W1 and W2

Now, to find the relationship between \(W_1\) and \(W_2\), we will divide the second equation by the first equation: \(\frac{W_2}{W_1} = \frac{(1/2)(3K)x^2}{(1/2)Kx^2}\)
03

Simplify the equation

Observe that \((1/2)x^2\) can be canceled out in both numerator and denominator, and we are left with: \(\frac{W_2}{W_1} = \frac{3K}{K}\)
04

Solve for the relationship between W1 and W2

Now, dividing both sides by K, we have \(\frac{W_2}{W_1} = 3\) Multiplying both sides by \(W_1\), we get: \(W_2 = 3\cdot W_1\)
05

Compare the result with the given options

Finally, we can compare our result to the given options. Our result matches option (D), which states that: \(W_2 = 3\cdot W_1\) Therefore, the correct answer is (D).

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Most popular questions from this chapter

If Wa, Wb, and Wc represent the work done in moving a particle from \(\mathrm{X}\) to \(\mathrm{Y}\) along three different path $\mathrm{a}, \mathrm{b}\(, and \)\mathrm{c}$ respectively (as shown) in the gravitational field of a point mass \(\mathrm{m}\), find the correct relation between \(\mathrm{Wa}, \mathrm{Wb}\) and \(\mathrm{Wc}\) (A) \(\mathrm{Wb}>\mathrm{Wa}>\mathrm{Wc}\) (B) \(\mathrm{Wa}<\mathrm{Wb}<\mathrm{Wc}\) (C) \(\mathrm{Wa}>\mathrm{Wb}>\mathrm{Wc}\) (D) \(\mathrm{Wa}=\mathrm{Wb}=\mathrm{Wc}\)

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 metal ball of mass \(2 \mathrm{~kg}\) moving with a velocity of $36 \mathrm{~km} / \mathrm{h}$ has a head on collision with a stationary ball of mass \(3 \mathrm{~kg}\). If after the collision, the two balls move together, the loss in kinetic energy due to collision is (A) \(40 \mathrm{~J}\) (B) \(60 \mathrm{~J}\) (C) \(100 \mathrm{~J}\) (D) \(140 \mathrm{~J}\)

A force \(\mathrm{F}=\mathrm{kx}\) (where \(\mathrm{k}\) is positive constant) is acting on a particle. Match column-I and column-II, regarding work done in displacing the particle. $$ \begin{array}{|l|l|} \hline \text { Column - i } & \text { Column - ii } \\ \hline \text { (a) From } \mathrm{x}=-4 \text { to } \mathrm{x}=-2 & \text { (P) Positive } \\ \hline \text { (b) From } \mathrm{x}=-2 \text { to } \mathrm{x}=-4 & \text { (Q) zero } \\ \hline \text { (c) From } \mathrm{x}=-2 \text { to } \mathrm{x}=+2 & \text { (R) negative } \\ \hline \end{array} $$ (A) \(\mathrm{a}-\mathrm{R}, \mathrm{b}-\mathrm{P}, \mathrm{c}-\mathrm{Q}\) (B) \(a-P, b-Q, c-R\) (C) \(a-R, b-Q, c-P\) (D) \(\mathrm{a}-\mathrm{Q}, \mathrm{b}-\mathrm{P}, \mathrm{c}-\mathrm{R}\)

A ball of mass \(5 \mathrm{~kg}\) is striding on a plane with initial velocity of \(10 \mathrm{~m} / \mathrm{s}\). If co-efficient of friction between surface and ball is \((1 / 2)\), then before stopping it will describe \(\ldots \ldots\) \(\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)\) (A) \(12.5 \mathrm{~m}\) (B) \(5 \mathrm{~m}\) (C) \(7.5 \mathrm{~m}\) (D) \(10 \mathrm{~m}\)
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