/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 32 A \(1 \mathrm{~kg}\) block moves... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 32

A \(1 \mathrm{~kg}\) block moves towards a light spring with a velocity of \(8 \mathrm{~m} / \mathrm{s}\). When the spring is compressed by \(3 \mathrm{~m}\), its momentum becomes half of the original momentum. Spring constant of the spring is (A) \(18 / 3 \mathrm{~N} / \mathrm{m}\) (B) \(16 / 3 \mathrm{~N} / \mathrm{m}\) (C) \(3 \mathrm{~N} / \mathrm{m}\) (D) \(8 \mathrm{~N} / \mathrm{m}\)

Short Answer

Expert verified
The spring constant is \(k = \frac{16}{3} N/m\), which corresponds to option (B).

Step by step solution

01

Calculate the initial momentum of the block.

To find the initial momentum of the block, we have to multiply its mass (m) by its initial velocity (v), which is: \(p_{initial} = m \cdot v\) Plugging in the given values, we have: \(p_{initial} = 1kg \cdot 8m/s = 8 kg \cdot m/s\)
02

Calculate the final momentum of the block.

The problem states that the final momentum is half of the initial momentum. So, we can find the final momentum: \(p_{final} = \frac{1}{2} \cdot p_{initial}\) \(p_{final} = \frac{1}{2} \cdot 8 kg \cdot m/s = 4 kg \cdot m/s\)
03

Calculate the change in kinetic energy.

Since the block's momentum changes, its kinetic energy also changes. We can find the change in kinetic energy using the following formula: \(\Delta KE = \frac{1}{2} m \cdot v_{final}^2 - \frac{1}{2} m \cdot v_{initial}^2\) We can find the final velocity (v_final) with the final momentum and mass: \(v_{final} = \frac{p_{final}}{m} = \frac{4 kg \cdot m/s}{1 kg} = 4 m/s\) Now, we can plug the values into the equation: \(\Delta KE = \frac{1}{2} \cdot 1kg \cdot (4 m/s)^2 - \frac{1}{2} \cdot 1kg \cdot (8 m/s)^2 = -24 J\)
04

Calculate the potential energy stored in the spring.

The change in kinetic energy is equal to the potential energy stored in the spring. So, we can write: \(\Delta KE = U_{spring}\) \(-24 J = \frac{1}{2} k \cdot x^2\) Here, k is the spring constant, and x is the compression of the spring (3 m).
05

Calculate the spring constant.

Rearrange the equation to find the spring constant (k): \(k = \frac{-2 \cdot U_{spring}}{x^2} = \frac{-2 \cdot (-24 J)}{(3 m)^2}\) \(k = \frac{48 J}{9 m^2} = \frac{16}{3} N/m\) The spring constant is \(k = \frac{16}{3} N/m\), which corresponds to option (B).

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Most popular questions from this chapter

A particle of mass \(m\) is attached to a light string of length \(l\), the other end of which is fixed. Initially, the string is kept horizontal and the particle is given an upward velocity \(v\). The particle is just able to complete a circle. (A) The string becomes slack when the particle reaches its highest point. (B) The velocity of the particle becomes zero at the highest point. (C) The kinetic energy of the ball in initial position was \(\frac{1}{2} m v^{2}=m g l\). (D) The particle again passes through the initial position.

An object of mass \(10 \mathrm{~kg}\) falls from rest through a vertical distance of \(10 \mathrm{~m}\) and acquires a velocity of \(10 \mathrm{~m} / \mathrm{s}\). The work done by the push of air on the object is \((g=\) \(\left.10 \mathrm{~m} / \mathrm{s}^{2}\right)\) (A) \(500 \mathrm{~J}\) (B) \(-500 \mathrm{~J}\) (C) \(250 \mathrm{~J}\) (D) \(-250 \mathrm{~J}\)

A particle of mass \(2 \mathrm{~kg}\) starts moving in a straight line with an initial velocity of \(2 \mathrm{~m} / \mathrm{s}\) at a constant acceleration of \(2 \mathrm{~m} / \mathrm{s}^{2}\). The rate of change of kinetic energy is (A) Four times the velocity at any moment. (B) Two times the displacement at any moment. (C) Four times the rate of change of velocity at any moment. (D) Constant throughout.

A cricket ball is hit for a six leaving the bat at an angle of \(45^{\circ}\) to the horizontal with kinetic energy \(K\). At the top position, the kinetic energy of the ball is (A) Zero (B) \(K\) (C) \(K / 2\) (D) \(K / \sqrt{2}\)

A block of mass \(m\) is slowly pulled up on inclined plane of height \(h\) and inclination \(\theta\) with the top of a string parallel to the incline. Which of the following statement is correct for the block when it moves up from the bottom to the top of the incline? (A) Work done by the normal reaction force is zero. (B) Work done by the string on block is \(m g h\). (C) Work done by the gravity is \(m g h\). (D) Work done by the block is \(-m g h / 2\).
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