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Chapter 4: Problem 6

A particle projected with an initial velocity \(u\) at angle \(\theta\) from the ground. The work done by gravity during the time it reaches the highest point \(P\) is: (A) \(\frac{-m u^{2} \sin ^{2} \theta}{2}\) (B) \(+\frac{m u^{2} \sin ^{2} \theta}{2}\) (C) 0 (D) \(+m u^{2} \sin \theta\)

Short Answer

Expert verified
The short answer, based on the step-by-step solution, is: The work done by gravity during the time the particle reaches the highest point P is \(\frac{-m u^{2} \sin ^{2} \theta}{2}\) (Answer A).

Step by step solution

01

Calculate the horizontal and vertical components of velocity

To solve this problem, we need to analyze the motion of the particle in two dimensions. The horizontal and vertical components of the initial velocity are given by: \(v_x = u \cos(胃)\) \(v_y = u \sin(胃)\)
02

Calculate the vertical velocity at the highest point

At the highest point P of the projectile's trajectory, the vertical component of the velocity goes to zero. Thus: \(v_{y_P} = 0\)
03

Calculate the change in kinetic energy

The initial and final kinetic energies of the particle are given by: \(KE_i = \frac{1}{2}m(u^2)\) (Initial kinetic energy) \(KE_P = \frac{1}{2}m(v_{x_P}^2)\) (Kinetic energy at the highest point P) The change in kinetic energy is given by the difference between the initial and final kinetic energies: 螖KE = \(KE_P - KE_i\)
04

Apply the work-energy theorem

According to the work-energy theorem, the work done by gravity is equal to the change in kinetic energy: W = 螖KE
05

Choose the correct answer based on the work-energy theorem

Substituting the expressions for KE_P and KE_i from Steps 3 and 4 into the previous equation for W, we get: W = \(\frac{1}{2}m(v_{x_P}^2 - u^2)\) Since \(v_{x_P}\) is just the horizontal component of the initial velocity, \(v_{x_P} = u \cos(胃)\) Substituting this into the equation for W, we get: W = \(\frac{1}{2}m(u^2\cos^2(胃) - u^2)\) This can be simplified to: W = \(\frac{-mu^2(\sin^2(胃))}{2}\) Comparing this result with the given answer choices, we conclude that Answer (A) is the correct answer: Work done by gravity = \(\frac{-m u^{2} \sin ^{2} \theta}{2}\)

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Most popular questions from this chapter

Two identical beads of \(m=100 \mathrm{~g}\) are connected by an inextensible massless string that can slide along the two arms \(A C\) and \(B C\) of a rigid smooth wireframe in a vertical plane. If the system is released from rest, the kinetic energy of the first particle when they have moved by a distance of \(0.1 \mathrm{~m}\) is \(16 x \times 10^{-3} \mathrm{~J}\). Find the value of \(x .\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)\)

A ball of mass \(m\) is attached to a light string of length \(L\) and suspended vertically. A constant horizontal force, whose magnitude \(F\) equals the weight of the ball is applied. The speed of the ball as it reaches \(90^{\circ}\) level is, (A) \(\sqrt{g L}\) (B) \(\sqrt{2 g L}\) (C) \(\sqrt{3 g L}\) (D) Zero

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