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Chapter 3: Problem 56

A block \(A\) of mass \(m\) is placed over a plank \(B\) of mass \(2 \mathrm{~m}\). Plank \(B\) is placed over a smooth horizontal surface. The co-efficient of friction between \(A\) and \(B\) is \(\frac{1}{2} .\) Block \(A\) is given a velocity \(v_{0}\) towards right. Acceleration of \(B\) relative to \(A\) is (A) \(\frac{g}{2}\) (B) \(g\) (C) \(\frac{3 g}{4}\) (D) Zero

Short Answer

Expert verified
The relative acceleration of plank B with respect to block A is not among the given options, but based on the calculated values and comparison, the correct answer is (D) Zero.

Step by step solution

01

Determine the forces acting on each object

For block A, there are only two forces acting: gravity (downward) and friction (in opposite direction of motion). Since there is no vertical motion, we only need to consider the frictional force: \[F_{friction} = \mu\cdot m_{A}g = \frac{1}{2} \cdot m g\] For plank B, there are no frictional forces acting on it directly, so the only force it experiences is the opposite reaction force from the friction between block A and plank B.
02

Calculate the acceleration of each object

Based on Newton's second law, we know that \(F=ma\). So, the acceleration of block A is: \[ a_{A} = \frac{F_{friction}}{m_{A}} = \frac{\frac{1}{2} \cdot m g}{m} = \frac{g}{2} \] And the acceleration of plank B is: \[ a_{B} = \frac{F_{friction}}{m_{B}} = \frac{\frac{1}{2} \cdot m g}{2m} = \frac{g}{4} \]
03

Calculate the relative acceleration of plank B with respect to block A

To find the relative acceleration, we subtract the acceleration of block A from the acceleration of plank B: \[ a_{B,rel} = a_{B} - a_{A} = \frac{g}{4} - \frac{g}{2} = -\frac{g}{4} \] The relative acceleration of plank B with respect to block A is \(-\frac{g}{4}\), which is not among the given options. However, since the acceleration of plank B is less than that of block A, it is clear that the right answer is (D), as the relative acceleration should be smaller than both \(\frac{g}{2}\) and \(g\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Newton's Second Law
Understanding Newton's second law is vital when examining the dynamics of objects. The law is succinctly defined by the equation \( F = ma \), where \( F \) stands for the force exerted on an object, \( m \) is the mass of the object, and \( a \) denotes its acceleration. In essence, this principle tells us that an object's acceleration is directly proportional to the net force acting upon it and inversely proportional to its mass. This relationship implies a larger force results in greater acceleration and that more massive objects will accelerate less than their lighter counterparts when subjected to the same amount of force.

For objects in contact, like a block sliding over a plank, Newton's second law helps us to calculate the individual accelerations of each object by considering the forces they exert on each other. In the problem provided, we examine how the frictional force, which acts as the only horizontal force on both block A and plank B, determines their accelerations and consequently their relative acceleration.
Frictional Force
Frictional force plays a pivotal role in this physics problem. It's a resistive force that occurs when two surfaces come into contact and move against each other. Friction is dependent not only on the nature of the surfaces but also on the force pressing them together. In the case at hand, the frictional force arises due to the movement between block A and plank B.

It's important to note that the frictional force always acts in the opposite direction to the movement of the objects, attempting to inhibit relative motion. For block A, friction is the only horizontal force acting, and according to Newton's second law, it's responsible for its acceleration. For plank B, which rests on a smooth surface with no friction, the frictional force comes from the contact with block A, causing it to accelerate in the opposite direction.
Coefficient of Friction
The coefficient of friction is a dimensionless scalar value which describes the ratio of the force of friction between two bodies and the force pressing them together. It's symbolized by \( \( \mu \ \) \) and varies based on the materials in contact and their surface properties. In our problem, the coefficient of friction between block A and plank B is given as \( \frac{1}{2} \).

The coefficient of friction is a key factor in calculating the frictional force. By multiplying it with the normal force (in this scenario, the weight of block A), we can determine the magnitude of the frictional force that acts to slow down block A and move plank B. It's crucial to remember that the coefficient of friction does not have a unit, as it is the ratio of two forces.

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Most popular questions from this chapter

A block of mass \(5 \mathrm{~kg}\) is kept on a rough horizontal floor. It's given velocity is \(33 \mathrm{~m} / \mathrm{s}\) towards right. A force of \(20 \sqrt{2} \mathrm{~N}\) continuously acts on the block as shown. If the co- efficient of friction between block and floor is \(0.5\), find the velocity of the block after 5 seconds \(\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)\).

A car starts from rest to cover a distance \(x\). The co-efficient of friction between the road and tyres is \(\mu\). The minimum time in which the car can cover distance \(x\) is proportional to (A) \(\mu\) (B) \(\frac{1}{\sqrt{\mu}}\) (C) \(\sqrt{\mu}\) (D) \(\frac{1}{\mu}\)

Starting from rest, a particle rotates in a circle of radius \(R=\sqrt{2} \mathrm{~m}\) with an angular acceleration \(\alpha=(\pi / 4) \mathrm{rad} / \mathrm{s}^{2} .\) The magnitude of average velocity of the particle over the time it rotates a quarter circle is (A) \(1.5 \mathrm{~m} / \mathrm{s}\) (B) \(2 \mathrm{~m} / \mathrm{s}\) (C) \(1 \mathrm{~m} / \mathrm{s}\) (D) \(1.25 \mathrm{~m} / \mathrm{s}\)

A motor car is traveling at \(60 \mathrm{~m} / \mathrm{s}\) on a circular road of radius \(1200 \mathrm{~m}\). It is increasing its speed at the rate of \(4 \mathrm{~m} / \mathrm{s}^{2}\). The acceleration of the car is (A) \(3 \mathrm{~ms}^{-2}\) (B) \(4 \mathrm{~ms}^{-2}\) (C) \(5 \mathrm{~ms}^{-2}\) (D) \(7 \mathrm{~ms}^{-2}\)

A block of mass \(m\) is connected to another block of mass \(M\) by a spring (massless) of spring constant \(k\). The blocks are kept on a smooth horizontal plane. Initially the blocks are at rest and the spring is unstretched. Then a constant force \(F\) starts acting on the block of mass \(M\) to pull it. Find the force on the block of mass \(m .\) (A) \(\frac{m F}{M}\) (B) \(\frac{(M+m) F}{m}\) (C) \(\frac{m F}{(m+M)}\) (D) \(\frac{M F}{(m+M)}\)
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