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Chapter 4: Problem 11

A particle is released from rest at origin. It moves under the influence of potential field \(U=x^{2}-3 x\), where \(U\) is in Joule and \(x\) is in metre. Kinetic energy at \(x=2 \mathrm{~m}\) will be (A) \(2 \mathrm{~J}\) (B) \(1 \mathrm{~J}\) (C) \(1.5 \mathrm{~J}\) (D) \(0 \mathrm{~J}\)

Short Answer

Expert verified
The correct answer is \(2 \mathrm{~J}\) (Option A). The kinetic energy at x = 2 meters is 2 J, which is calculated using conservation of energy and the given potential field formula.

Step by step solution

01

Determine initial and final potential energy

We are given the potential field U = x^2 - 3x. To find the initial and final potential energies, we will plug in the initial position x = 0 meters and the final position x = 2 meters to get the initial and final potential energies. Initial potential energy: U_initial = (0)^2 - 3(0) = 0 J Final potential energy: U_final = (2)^2 - 3(2) = 4 - 6 = -2 J
02

Calculate the change in potential energy

Next, we need to find the change in potential energy. To do this, subtract the initial potential energy from the final potential energy (U_final - U_initial). Change in potential energy: 螖U = U_final - U_initial = (-2 J) - (0 J) = -2 J
03

Use conservation of energy to find the final kinetic energy

Now that we have the change in potential energy, we can use conservation of energy to find the final kinetic energy of the particle. The conservation of energy states that the sum of kinetic and potential energies will remain constant. Since the particle is initially at rest, its initial kinetic energy is 0 J. So, we can write the conservation of energy equation as: K_initial + U_initial = K_final + U_final Substituting the known values: 0 J + 0 J = K_final - 2 J
04

Solve for the final kinetic energy

From the conservation of energy equation, we can now solve for the final kinetic energy: K_final = 0 J + 2 J = 2 J The final kinetic energy at x = 2 meters is 2 J. Therefore, the correct answer is (A) 2 J.

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Most popular questions from this chapter

A car comes to a skidding stop in \(15 \mathrm{~m}\). The force on the car due to the road is \(1000 \mathrm{~N}\). The work done by road on the car and car on the road, respectively, is (A) \(-15 \mathrm{~kJ}\), zero (B) zero, \(15 \mathrm{~kJ}\) (C) \(15 \mathrm{~kJ}\), zero (D) \(-15 \mathrm{~kJ}, 15 \mathrm{~kJ}\)

A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle. The motion of the particle takes place in a plane. It follows that (A) its velocity is constant. (B) its acceleration is constant. (B) its kinetic energy is constant. (D) it moves in a circular path.

A \(2 \mathrm{~kg}\) block slides on a horizontal floor with a speed of \(4 \mathrm{~m} / \mathrm{s}\). It strikes an uncompressed spring, and compresses it till the block is motionless. The kinetic friction force is \(15 \mathrm{~N}\) and spring constant is \(10,000 \mathrm{~N} / \mathrm{m}\). The spring compresses by \([\mathbf{2 0 0 7}]\) (A) \(8.5 \mathrm{~cm}\) (B) \(5.5 \mathrm{~cm}\) (C) \(2.5 \mathrm{~cm}\) (D) \(11.0 \mathrm{~cm}\)

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