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The coefficient of friction between the block and plank is \(\mu\) and its value is such that the block becomes stationary with respect to plank before it reaches the other end. Then: (1) The work done by friction on the block is negative. (2) The work done by friction on the plank is positive. (3) The net work done by friction is negative. (4) Net work done by the friction is zero.

Short Answer

Expert verified
(1), (2), and (4) are correct; net work by friction is zero.

Step by step solution

01

Analyze the System

We have a block on a plank. Since the block becomes stationary with respect to the plank due to friction, the frictional force must be acting in such a way that it counters the motion of the block relative to the plank.
02

Determine the Work Done by Friction on the Block

Friction acts in the opposite direction to the block's motion, therefore, it does negative work on the block as it slows it down relative to the plank. Hence, statement (1) is correct.
03

Determine the Work Done by Friction on the Plank

The friction exerts force on the plank in the direction of the block's initial motion. This contributes to the motion of the plank in that direction, meaning the work done by friction on the plank is positive. Hence, statement (2) is correct.
04

Calculate the Net Work Done by Friction

The work done by friction on the block and the plank is equal and opposite, as they act as action and reaction pairs. Thus, the net work done by the frictional forces (block and plank combined) sums to zero. Hence, statement (4) is correct and statement (3) is incorrect.

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Key Concepts

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

Work Done by Friction
Friction plays a vital role in the interaction between surfaces because it opposes the relative motion between them. The work done by friction on an object is calculated as \[ W = F_f \cdot d \cdot \cos(\theta) \]where:
  • \( W \) is the work done,
  • \( F_f \) is the frictional force,
  • \( d \) is the distance over which the force acts,
  • \( \theta \) is the angle between the force and the direction of motion.
In a block and plank system, if the friction is working against an object's motion, the angle involved would generally be 180 degrees, making \( \cos(\theta) = -1 \). This results in negative work when friction slows down or stops the object. On the other hand, if friction aids in moving another body, like the plank in the specified direction of the block's initial motion, it can perform positive work. This dual effect is an excellent example of how friction can act differently depending on the situation.
Block and Plank System
A block and plank system is a simple model widely used to understand friction and motion. In this system, the block is placed on the plank, and both can move relative to each other or together. The frictional force between the surfaces is central to predicting how the block and plank will move. When the block initially moves relative to the plank, friction acts to slow it down. As friction acts in opposition to the block's motion, it eventually comes to a stop and moves in unison with the plank. For the plank, however, friction provides a force in the direction of the block’s initial motion, which could potentially move the plank further. This relationship is key to solving problems involving friction, as it demonstrates how the forces work in tandem to influence motion.
  • The block slows and stops due to negative work from friction.
  • The plank receives a push from friction, allowing it to move, illustrating positive work done on it.
Understanding these nuances aids in predicting motion in similar systems.
Newton's Third Law
Newton's Third Law is famously summarized as: "For every action, there is an equal and opposite reaction." This principle is pivotal in systems involving friction, such as the block and plank system, where forces are at play in an interactive manner. When the block experiences frictional force that acts opposite to its direction of motion, the plank simultaneously experiences a force of equal magnitude in the opposite direction. This reciprocal action explains why the work done by friction on the block and the work done on the plank can have opposite signs. In our case:
  • The block experiences negative work due to friction resisting its motion.
  • The plank experiences positive work because the same force assists its motion.
  • Ultimately, these opposite works balance each other out, demonstrating Newton's Third Law in action.
By internalizing these dynamics, one can better analyze and solve problems that involve pairs of action and reaction forces in similar real-world scenarios.

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Most popular questions from this chapter

A block of mass \(m\) lies on a wedge of mass \(M\). The wedge in tum lies on a smooth horizontal surface. Friction is absent everywhere. The wedge-block system is released from rest. All situations given in Column I are to be estimated in duration the block undergoes a vertical displacement \(h\) starting from rest. Match the statements in Column I with the results in Column II. ( \(g\) is acceleration due to gravity.) $$ \begin{array}{|c|c|} \hline {\begin{array}{c} \text { Column I } \\ \end{array}} & {\begin{array}{c} \text { Column II } \\ \end{array}} \\ \hline \text { i. } \text {Work done by normal reaction acting on the block is} & \text { a. } \text{positive} \\ \hline \text { ii. } \text{Work done by normal reaction (exerted by block) acting on the wedge is} & \text { b. } \text{negative} \\ \hline \text { iii. } \text{The sum of work done by normal reaction on the block and work done by normal on wedge} & \text { c. } \text{Zero}\\\ \hline \text { iv. } \text{Net work done by all forces on the block is} & \text { d. } \text{less than \(m g h\) in magnitude} \\ \hline \end{array} $$

The block of mass \(m\) initially at \(x=0\) is acted upon by a horizontal force \(F=a-b x^{2}\) (where \(a>\mu m g\) ), as shown in the figure. The co-efficient of friction between the surfaces of contact is \(\mu\). The net work done on the block is zero, if the block travels a distance of (1) \(x=\sqrt{\frac{3(a-\mu m g)}{b}}\) (2) \(x=\sqrt{\frac{3(a+\mu m g)}{b}}\) (3) \(x=\sqrt{\frac{2(a-\mu m g)}{b}}\) (4) \(x=\sqrt{\frac{2(a+\mu m g)}{b}}\)

Two springs \(P\) and \(Q\) having stiffness constants \(k_{1}\) and \(k_{2}\left(

A chain of length \(l\) and mass \(m\) lies on the surface of a smooth sphere of radius \(R>l\) with one end tied to the top of the sphere. $$ \begin{array}{|c|c|} \hline {\begin{array}{c} \text { Column I } \\ \end{array}} & {\begin{array}{c} \text { Column II } \\ \end{array}} \\ \hline \text { i. } \text {Gravitational potential energy w.r.t. centre of the sphere} & \text { a. } \text{\(\frac{\operatorname{Rg}}{l}\left[1-\cos \left(\frac{l}{R}\right)\right]\)} \\ \hline \text { ii. } \text{The chain is released and slides down, its \(\mathrm{KE}\) when it has slid by \(\theta\)} & \text { b. } \text{\(\frac{2 R g}{l}\left[\sin \left(\frac{l}{R}\right)+\sin \theta-\sin \left(\theta+\frac{l}{R}\right)\right]\)} \\ \hline \text { iii. } \text{The initial tangential acceleration} & \text { c. } \text{\(\frac{M R^{2} g}{l} \sin \left(\frac{l}{R}\right)\)}\\\ \hline \text { iv. } \text{The radial acceleration \(a_{r}\)} & \text { d. } \text{\(\frac{M R^{2} g}{l}\left[\sin \left(\frac{l}{R}\right)+\sin \theta-\sin \left(\theta+\frac{l}{R}\right)\right]\)} \\ \hline \end{array} $$

A ball of mass \(m\) is attached to the lower end of light vertical spring of force constant \(k\). The upper end of the spring is fixed. The ball is released from rest with the spring at its normal (unstretched) length and comes to rest again after descending through a distance \(x\). (1) \(x=m g / k\) (2) \(x=2 m g / k\) (3) The ball will have no acceleration at the position where it has descended through \(x / 2\). (4) The ball will have an upward acceleration equal to \(g\) at its lowermost position

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