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Chapter 4: Problem 442

A particle of mass \(0.5 \mathrm{~kg}\) travels in a straight line with velocity \(\mathrm{v}=\mathrm{ax}^{3 / 2}\), Where $\mathrm{a}=5 \mathrm{~m}^{[(-1) / 2]} \mathrm{s}^{-1}$. The work done by the net force during its displacement from \(\mathrm{x}=0\) to $\mathrm{x}=2 \mathrm{~m}$ is (A) \(50 \mathrm{~J}\) (B) \(45 \mathrm{~J}\) (C) \(25 \mathrm{~J}\) (D) None of these

Short Answer

Expert verified
The work done by the net force during the particle's displacement from \(x = 0 \mathrm{~m}\) to \(x = 2 \mathrm{~m}\) is \(50 \mathrm{~J}\). The correct answer is (A) \(50 \mathrm{~J}\).

Step by step solution

01

Find the initial and final velocities

First, evaluate the velocity function at the initial and final displacements: \(x = 0 \mathrm{~m}\) and \(x = 2 \mathrm{~m}\). Initial velocity (\(v_0\)): \[v_0 = a x^{3/2} = 5(0)^{3/2} = 0 \mathrm{~m/s}\] Final velocity (\(v_f\)): \[v_f = a x^{3/2} = 5(2)^{3/2} = 5(2\sqrt{2}) = 10\sqrt{2} \mathrm{~m/s}\]
02

Calculate the initial and final kinetic energies

Using the mass of the particle, we can calculate its initial and final kinetic energies, given by the formula \(K = \frac{1}{2}mv^2\). Initial kinetic energy (\(K_0\)): \[K_0 = \frac{1}{2}mv_0^2 = \frac{1}{2}(0.5 \mathrm{~kg})(0 \mathrm{~m/s})^2 = 0 \mathrm{~J}\] Final kinetic energy (\(K_f\)): \[K_f = \frac{1}{2}mv_f^2 = \frac{1}{2}(0.5 \mathrm{~kg})(10\sqrt{2} \mathrm{~m/s})^2 = 50 \mathrm{~J}\]
03

Calculate the work done using the work-energy theorem

According to the work-energy theorem, the work done (\(W\)) is equal to the change in kinetic energy, which is the difference between the final and initial kinetic energies: \[W = K_f - K_0 = 50 \mathrm{~J} - 0 \mathrm{~J} = 50 \mathrm{~J}\] Thus, the work done by the net force during the particle's displacement from \(x = 0 \mathrm{~m}\) to \(x = 2 \mathrm{~m}\) is \(50 \mathrm{~J}\). The correct answer is (A) \(50 \mathrm{~J}\).

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