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Chapter 8: Problem 1

When a body is moving vertically up with constant velocity, then match the following. $$ \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 lifting force is} & \text { a. } \text{negative} \\ \hline \text { ii. } \text{Total work done by all the forces is} & \text { b. } \text{positive} \\ \hline \text { iii. } \text{Work done by gravity} & \text { c. } \text{zero}\\\ \hline \text { iv. } \text{Work done by lifting force \(+\) work done by gravity force} & \text { d. } \text{higher positive values} \\ \hline \end{array} $$

Short Answer

Expert verified
i-b, ii-c, iii-a, iv-c.

Step by step solution

01

Understanding the problem

The problem involves matching statements about work done in a vertical motion scenario with the corresponding qualitative outcomes from a given list. The body moves with constant velocity, implying a balance of forces.
02

Lifting force work

When a body moves with constant velocity vertically, the lifting force is doing positive work because it is moving the body against gravity at a steady rate. Match 'Work done by lifting force is' with 'positive.'
03

Total work done by all forces

In constant velocity motion, the net force is zero, meaning no acceleration. Hence, total work done is zero, not causing any change in kinetic energy. Match 'Total work done by all the forces is' with 'zero.'
04

Work done by gravity

Gravity always does negative work when moving upwards as it acts in the opposite direction to the motion. Match 'Work done by gravity' with 'negative.'
05

Sum of work done by lifting and gravity

The algebraic sum of work done by lifting force and gravity force is zero due to constant velocity (net force is zero). Match 'Work done by lifting force + work done by gravity force' with 'zero.'

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Key Concepts

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

Constant Velocity
When a body moves vertically with a constant velocity, it means there's no change in its speed or direction over time. This scenario is key in understanding forces and work done on the object. To maintain constant velocity, the sum of forces must be balanced. Therefore, the forces acting on the body do not accelerate it. Instead, they counteract each other perfectly.

* Forces in constant velocity include: - **Lifting Force:** This is the force applied to lift the object upward. - **Gravitational Force:** This force pulls the object back towards Earth. Together, these forces ensure that the object does not speed up or slow down. Since the forces are in equilibrium, the net external force acting on the object is zero, which aligns with Newton's First Law of Motion.
Work Done by Gravity
Work is done by a force when it causes displacement. For an object moving upward with constant velocity, gravity acts downward. Consequently, the work done by gravity is considered negative because it opposes the direction of displacement of the object.

* Key points about work done by gravity: - The direction of gravitational force is opposite to the direction of motion, hence the work is negative. - The magnitude of work done by gravity is equal to the weight of the body multiplied by the distance moved. - Formula: Work done by gravity = \[ W_{gravity} = -mgd \]Where:- \( m \) is the mass of the object,- \( g \) is the acceleration due to gravity,- \( d \) is the vertical displacement.Understanding this concept is crucial in grasping how forces interact during vertical motion and calculating the net work done.
Lifting Force
The lifting force is crucial when an object is moving upwards at constant velocity. For the body to maintain this constant speed, the lifting force applied must exactly counteract the force of gravity. Thus, the work done by the lifting force is positive.

* Characteristics of work done by lifting force: - Positive work is done, as it moves the object in the direction of force. - The force must be equal in magnitude to the gravitational force, ensuring equilibrium.The equation for work done by the lifting force is:\[W_{lifting} = mgd\]Where:- \( m \) is the mass of the object,- \( g \) is the gravitational pull,- \( d \) is the distance over which the force is applied.As we calculate the work done by lifting force and gravity, their sum invariably results in zero 鈥� illustrating the work-energy principle in cases of constant velocity.

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Most popular questions from this chapter

A man is standing on a plank which is placed on smooth horizontal surface. There is sufficient friction between feet of man and plank. Now man starts running over plank. The correct statements is/are (1) Work done by friction on man with respect to ground is negative (2) Work done by friction on man with respect to ground is positive (3) Work done by friction on plank with respect to ground is positive (4) Work done by friction on man with respect to plank is zero

A \(1.5-\mathrm{kg}\) block is initially at rest on a horizontal frictionless surface when a horizontal force in the positive direction of \(x\)-axis is applied to the block. The force is given by \(\vec{F}=\left(4-x^{2}\right) \vec{i} \mathrm{~N}\), where \(x\) is in meter and the initial position of the block is \(x=0\). The maximum positive displacement \(x\) is (1) \(2 \sqrt{3} \mathrm{~m}\) (2) \(2 \mathrm{~m}\) (3) \(4 \mathrm{~m}\) (4) \(\sqrt{2} \mathrm{~m}\)

An object of mass \(m\) slides down a hill of arbitrary shape and after travelling a certain horizontal path stops because of friction. The total vertical height descended is \(h\). The friction coefficient is different for different segments for the entire path but is independent of the velocity and direction of motion. The work that a tangential force must perform to return the object to its initial position along the same path is (1) \(\mathrm{mgh}\) (2) \(-m g h\) (3) \(-2 m g h\) (4) \(2 \mathrm{mgh}\)

The potential energy of a particle is determined by the expression \(U=\alpha\left(x^{2}+y^{2}\right)\), where \(\alpha\) is a positive constant. The particle begins to move from a point with coordinates \((3,3)\), only under the action of potential field force. Then its kinetic energy \(T\) at the instant when the particle is at a point with the coordinates \((1,1)\) is (1) \(8 \alpha\) (2) \(24 \alpha\) (3) \(16 \alpha\) (4) Zero

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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