An object of mass, m is moving with a constant velocity, v. How much work should be done on the object in order to bring the object to rest?
Approach Solution - 1
When an object is moving with a constant velocity, its kinetic energy is given by the equation:
Kinetic Energy= \(\frac 12mv^2\)
If you want to bring the object to rest, you need to remove all of its kinetic energy. The work done (W) is equal to the change in kinetic energy. In this case, the change in kinetic energy is equal to the initial kinetic energy (when the object is moving with velocity v).
So, the work done (W) to bring the object to rest is given by:
W= Change in Kinetic Energy = Final Kinetic Energy − Initial Kinetic Energy
Since the final kinetic energy is zero (the object is at rest), and the initial kinetic energy is \(\frac 12mv^2\), the work done is:
\(W=0−\frac 12mv^2\)
Therefore, the work done to bring the object to rest is \(-\frac 12mv^2\). The negative sign indicates that work is done against the motion of the object, removing its kinetic energy.
Approach Solution -2
When an object is brought to rest from a constant velocity, work must be done to overcome the object's kinetic energy and reduce it to zero. The work done W to bring an object of mass m moving with a constant velocity v to rest is equal to the change in kinetic energy.
The kinetic energy KE is given by the formula:
\(KE = \frac{1}{2} m v^2\)
When the object comes to rest, its final velocity \(v_f\) is 0, so the change in kinetic energy \(\Delta KE\) is:
\(\Delta KE = KE_{\text{final}} - KE_{\text{initial}}\)
\(\Delta KE = 0 - \frac{1}{2} m v^2\)
\(\Delta KE = -\frac{1}{2} m v^2\)
Since work is the transfer of energy, the work done on the object is equal to the change in kinetic energy:
\(W = \Delta KE\)
\(W = -\frac{1}{2} m v^2\)
So, the work done on the object to bring it to rest is \(-\frac{1}{2} m v^2\) or \(\frac{-1}{2}\) times its initial kinetic energy. The negative sign indicates that the work done is in the opposite direction of the object's initial motion.
So, the answer is: \(-\frac{1}{2} m v^2\)
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Questions Asked in CBSE Class IX exam
- Fig shows the distance-time graph of three objects A, B and C. Study the graph and answer the following questions:
- Which of the three is travelling the fastest?
- Are all three ever at the same point on the road?
- How far has C travelled when B passes A?
- How far has B travelled by the time it passes C?
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- Graphical Representation of Motion
- A driver of a car travelling at \(52\) \(km \;h^{–1}\) applies the brakes Shade the area on the graph that represents the distance travelled by the car during the period. (b) Which part of the graph represents uniform motion of the car?
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- Graphical Representation of Motion
- Name the following.
- Tissue that forms the inner lining of our mouth.
- Tissue that connects muscle to bone in humans.
- Tissue that transports food in plants.
- Tissue that stores fat in our body.
- Connective tissue with a fluid matrix.
- Tissue present in the brain.
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- Animal Tissues
Use these adverbs to fill in the blanks in the sentences below.
awfully sorrowfully completely loftily carefully differently quickly nonchalantly(i) The report must be read ________ so that performance can be improved.
(ii) At the interview, Sameer answered our questions _________, shrugging his shoulders.
(iii) We all behave _________ when we are tired or hungry.
(iv) The teacher shook her head ________ when Ravi lied to her.
(v) I ________ forgot about it.
(vi) When I complimented Revathi on her success, she just smiled ________ and turned away.
(vii) The President of the Company is ________ busy and will not be able to meet you.
(viii) I finished my work ________ so that I could go out to play
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- The fun they had
(Street Plan) : A city has two main roads which cross each other at the centre of the city. These two roads are along the North-South direction and East-West direction.
All the other streets of the city run parallel to these roads and are 200 m apart. There are 5 streets in each direction. Using 1cm = 200 m, draw a model of the city on your notebook. Represent the roads/streets by single lines. There are many cross- streets in your model. A particular cross-street is made by two streets, one running in the North - South direction and another in the East - West direction. Each cross street is referred to in the following manner : If the 2nd street running in the North - South direction and 5th in the East - West direction meet at some crossing, then we will call this cross-street (2, 5). Using this convention, find:
(i) how many cross - streets can be referred to as (4, 3).
(ii) how many cross - streets can be referred to as (3, 4).
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- Coordinate Geometry
Concepts Used:
Work
Work is the product of the component of the force in the direction of the displacement and the magnitude of this displacement.
Work Formula:
W = Force × Distance
Where,
Work (W) is equal to the force (f) time the distance.
Work Equations:
W = F d Cos θ
Where,
W = Amount of work, F = Vector of force, D = Magnitude of displacement, and θ = Angle between the vector of force and vector of displacement.
Unit of Work:
The SI unit for the work is the joule (J), and it is defined as the work done by a force of 1 Newton in moving an object for a distance of one unit meter in the direction of the force.
Work formula is used to measure the amount of work done, force, or displacement in any maths or real-life problem. It is written as in Newton meter or Nm.