/*! 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 45 A block of mass \(m\) is release... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 45

A block of mass \(m\) is released from rest when the extension in the spring is \(x_{0}\). The maximum downward displacement of the block is (A) \(\frac{m g}{2 k}-x_{0}\) (B) \(\frac{m g}{2 k}+x_{0}\) (C) \(\frac{2 m g}{k}-x_{0}\) (D) \(\frac{2 m g}{k}+x_{0}\)

Short Answer

Expert verified
The maximum downward displacement of the block is \(\frac{mg}{2k} + x_0\), which corresponds to the answer choice (B).

Step by step solution

01

Write down the potential energy at the initial position

The block is initially at a position \(x_0\) from the equilibrium position. The potential energy of the spring at this position is given by the formula \(U_s = \frac{1}{2} k x_0^2\).
02

Find the mechanical energy at the equilibrium position

When the block is at its equilibrium position, the spring is neither compressed nor extended, thus the potential energy of the spring is zero. The only energy present in the system is due to the gravitational force acting on the block. The gravitational potential energy is given by the formula \(U_g = mgh\), where \(h = x_0\) in this case because the block moves downward by the displacement \(x_0\). Therefore, the total mechanical energy at the equilibrium position is \(E = U_g = mgx_0\).
03

Write an expression for the total mechanical energy at the maximum displacement

At the maximum displacement, the total mechanical energy is the sum of the potential energy stored in the spring and the gravitational potential energy. Let the maximum displacement be \(x_m\). The potential energy of the spring is given by \(U_s = \frac{1}{2} k (x_m - x_0)^2\) and the gravitational potential energy is given by \(U_g = mgx_m\). Therefore, the total mechanical energy at the maximum displacement is \(E = \frac{1}{2} k (x_m - x_0)^2 + mgx_m\).
04

Equate the total mechanical energy at both positions and solve for the maximum displacement

Since the total mechanical energy is conserved, we equate the total mechanical energy at the initial position and at the maximum displacement. \(mgx_0 = \frac{1}{2} k (x_m - x_0)^2 + mgx_m\) Now, solve for \(x_m\): \(x_m = \frac{mg}{k} + x_0 - \frac{mg}{2k}\) Simplifying the expression, we get: \(x_m = \frac{2mg}{k} -\frac{mg}{2k} + x_0 = \frac{mg}{2k} + x_0\) So, the maximum downward displacement of the block is given by \(\frac{mg}{2k} + x_0\), which corresponds to the answer choice (B).

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Most popular questions from this chapter

A block of mass \(m\) is pulled by a constant power \(P\) placed on a rough horizontal plane. The friction co-efficient between the block and surface varies with its speed \(v\) as \(\mu=\frac{1}{\sqrt{1+v}} .\) The acceleration of the block when its speed is \(3 \mathrm{~m} / \mathrm{s}\) will be (A) \(\frac{P}{3 m}-\frac{g}{2}\) (B) \(\frac{P}{3 m}+\frac{g}{2}\) (C) \(\frac{P}{3 m}\) (D) \(\frac{g}{2}\)

When a rubber band is stretched by a distance \(x\), it exerts a restoring force of magnitude \(F=a x+\) \(b x^{2}\), where \(a\) and \(b\) are constants. The work done in stretching the unstretched rubber band by \(L\) is: [2014] (A) \(a L^{2}+b L^{3}\) (B) \(\frac{1}{2}\left(a L^{2}+b L^{3}\right)\) (C) \(\frac{a L^{2}}{2}+\frac{b L^{3}}{3}\) (D) \(\frac{1}{2}\left(\frac{2 L^{2}}{2}+\frac{b L^{2}}{3}\right)\)

A block of mass \(m=0.1 \mathrm{~kg}\) is released from a height of \(4 \mathrm{~m}\) on a curved smooth surface. On the horizontal surface, path \(A B\) is smooth and path \(B C\) offers coefficient of friction \(\mu=0.1\). If the impact of block with the vertical wall at \(C\) be perfectly elastic, the total distance covered by the block on the horizontal surface before coming to rest will be: (take \(g=10 \mathrm{~m} / \mathrm{s}^{2}\) ) (A) \(29 \mathrm{~m}\) (B) \(49 \mathrm{~m}\) (C) \(59 \mathrm{~m}\) (D) \(109 \mathrm{~m}\)

Two blocks \(A\) and \(B\) of equal mass \(m=1 \mathrm{~kg}\) are lying on a smooth horizontal surface as shown in Fig. \(4.27\). A spring of force constant \(K=200 \mathrm{~N} / \mathrm{m}\) is fixed at one end of block \(A\). Block \(B\) collides with another end of the spring with velocity \(v_{0}=2 \mathrm{~m} / \mathrm{s}\). What will be the maximum compression of the spring? [in decimeter \(]\)

Figure \(4.22\) shows a massless spring fixed at the bottom end of an inclined of inclination \(37^{\circ}\left(\tan 37^{\circ}=\right.\) \(3 / 4\) ). A small block of mass \(2 \mathrm{~kg}\) start slipping down the incline from a point \(4.8 \mathrm{~m}\) away from free end of spring. The block compresses the spring by \(20 \mathrm{~cm}\), stops momentarily and then rebounds through a distance \(1 \mathrm{~m}\) up the inclined, then \(\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)\). (A) Coefficient of friction between block and inclined is \(0.5\). (B) Coefficient of friction between block and inclined is \(0.75\). (C) Value of spring constant is \(1000 \mathrm{~N} / \mathrm{m}\). 69 (D) Value of spring constant is \(2000 \mathrm{~N} / \mathrm{m}\).
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