/*! 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 46 A small block of mass \(m\) lyin... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 46

A small block of mass \(m\) lying at rest at point \(P\) of a wedge having a smooth semi-circular track of radius \(R\). What should be the minimum value of horizontal acceleration \(a_{0}\) of wedge so that mass can just reach the point \(Q\) ? (A) \(g / 2\) (B) \(\sqrt{g}\) (C) \(g\) (D) Not possible

Short Answer

Expert verified
The correct answer is (D) Not possible, as the minimum horizontal acceleration required for the mass to reach point \(Q\) is \(\frac{g\sqrt{3}}{2}\), which doesn't match any of the provided options.

Step by step solution

01

Find the centripetal acceleration

The centripetal acceleration required for the mass to move in a circle of radius \(R\) can be found using the following equation: \[a_{c} = \frac{v^2}{R}\] Where \(v\) is the tangential velocity.
02

Balance the forces acting on the mass

When the mass is in equilibrium, the components of gravitational force acting upon it must be balanced by other forces. First, let's find the vertical component of gravitational force. The angle between the line connecting the mass to the center of the circle (radius) and vertical axis is \(90掳\), so the vertical force acting on the mass is: \[F_{vertical} = m\frac{g}{2}\] And, the horizontal component of gravitational force acting on the mass is: \[F_{horizontal} = mg\frac{\sqrt{3}}{2}\]
03

Calculate the horizontal acceleration to balance the forces

To balance the horizontal component of gravitational force acting on the mass, the wedge must exert an equal and opposite force. Therefore, the horizontal acceleration \(a_0\) must be equal to the horizontal force acting on the mass. \[ma_0 = F_{horizontal}\] \[a_0 = \frac{mg\sqrt{3}}{2m}\] Therefore, the minimum value of horizontal acceleration required for the mass to reach point \(Q\) is: \[a_0 = \frac{g\sqrt{3}}{2}\] None of the provided options (A, B, C, or D) match this result, so the correct answer is: (D) Not possible

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Most popular questions from this chapter

A body is moved along a straight line by a machine delivering a constant power. The distance moved by the body in time \(t\) is proportional to (A) \(t^{3 / 4}\) (B) \(t^{3 / 2}\) (C) \(t^{1 / 4}\) (D) \(t^{1 / 2}\)

One end of a light spring of spring constant \(k\) is fixed to a wall and the other end is tied to a block placed on a smooth horizontal surface. In a displacement, the work done by the spring is \(\frac{1}{2} k x^{2}\). The possible cases are: (A) The spring was initially compressed by a distance \(x\) and was finally in its natural length. (B) It was initially stretched by a distance \(x\) and finally was in its natural length. (C) It was initially in its natural length and finally in a compressed position. (D) It was initially in its natural length and finally in a stretched position.

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) \(109 \mathrm{~m}\)

A \(1 \mathrm{~kg}\) block collides with a horizontal light spring of force constant \(2 \mathrm{~N} / \mathrm{m}\). The maximum compression in the spring is \(4 \mathrm{~m}\). Assuming that co-efficient of kinetic friction between the block and the horizontal surface is \(0.25\), what is initial speed of the block (approx.)?

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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