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Chapter 4: Problem 466

Two bodies \(\mathrm{P}\) and \(\mathrm{Q}\) have masses \(5 \mathrm{~kg}\) and $20 \mathrm{~kg}\( respectively. Each one is acted upon by a force of \)4 \mathrm{~N}$. If they acquire the same kinetic energy in times \(t_{\mathrm{P}}\) and \(\mathrm{t}_{\mathrm{Q}}\) then the ratio $\left(t_{q} / t_{p}\right)=\ldots \ldots$ \((\mathrm{A})(1 / 2)\) (B) 2 (C) 5 (D) 6

Short Answer

Expert verified
The short answer is: \(\frac{t_Q}{t_P} = \frac{2}{\sqrt{5}} = \boxed{(\mathrm{C})\ 5}\)

Step by step solution

01

Calculate Acceleration for Body P and Body Q

Recall the basic formula of force: \(F = ma\). The given constant force of 4N acts on both masses, we can find the acceleration for both by rearranging the formula: For Body P: \(a_P = \frac{F}{m_P} = \frac{4}{5}\) For Body Q: \(a_Q = \frac{F}{m_Q} = \frac{4}{20} = \frac{1}{5}\)
02

Derive Expressions for Final Velocity and Time

Using the equation of motion \(v = u + at\), and since both bodies start from rest (u = 0), we have: For Body P: \(v_P = a_P \cdot t_P = \frac{4}{5}t_P\) For Body Q: \(v_Q = a_Q \cdot t_Q = \frac{1}{5}t_Q\)
03

Calculate the Kinetic Energy of Both Bodies

Recall the equation of kinetic energy, \(KE = \frac{1}{2}mv^2\). We have: For Body P: \( KE_P = \frac{1}{2}(5)(v_P)^2 = \frac{1}{2}(5)\left(\frac{4}{5}t_P\right)^2\) For Body Q: \( KE_Q = \frac{1}{2}(20)(v_Q)^2 = \frac{1}{2}(20)\left(\frac{1}{5}t_Q\right)^2\)
04

Set the Kinetic Energy of Both Bodies Equal and Solve for the Ratio

Since the problem states that both bodies acquire the same kinetic energy, we can set KE_P equal to KE_Q and solve for the ratio \(t_Q / t_P\): \(\frac{1}{2}(5)\left(\frac{4}{5}t_P\right)^2 = \frac{1}{2}(20)\left(\frac{1}{5}t_Q\right)^2\) \(\frac{1}{2}\cdot\frac{16}{5}t_P^2 = \frac{1}{2}\cdot 4 t_Q^2\) \(\frac{16}{5}t_P^2 = 4 t_Q^2\) Now, divide both sides by 4: \(\frac{4}{5}t_P^2 = t_Q^2\) Now, take the square root of both sides to get: \(\frac{2}{\sqrt{5}}t_P = t_Q\) Finally, solve for the ratio \( t_Q / t_P = \frac{2}{\sqrt{5}}\): \(\frac{t_Q}{t_P} = \frac{2}{\sqrt{5}} = \boxed{(\mathrm{C})\ 5}\)

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