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: \(-\frac{1}{2} m v^2\)
Top Questions on Work
- In an actual regenerative Rankine cycle, the net work out put will be _____ as compared to simple Ranking cycle.
- A metal has work function \(w_0\) = 3.3×10-19 J. What should be the minimum frequency of the incident radiation that can remove an electron from the metal surface? [Given h = 6.6x10-34 J-s]
- In each of the following a force, F is acting on an object of mass, m. The direction of displacement is from west to east shown by the longer arrow. Observe the diagrams carefully and state whether the work done by the force is negative, positive or zero.
- Look at the activities listed below. Reason out whether or not work is done in the light of your understanding of the term ‘work’.
• Suma is swimming in a pond.
• A donkey is carrying a load on its back.
• A wind-mill is lifting water from a well.
• A green plant is carrying out photosynthesis.
• An engine is pulling a train.
• Food grains are getting dried in the sun.
• A sailboat is moving due to wind energy. - Calculate the work required to be done to stop a car of 1500 kg moving at a velocity of 60 km/h?
Questions Asked in CBSE Class IX exam
Factorise each of the following:
(i) 27y 3 + 125z 3
(ii) 64m3 – 343n 3
[ Hint : See Question 9. ]
- CBSE Class IX
- Algebraic Identities
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
- CBSE Class IX
- The fun they had
- 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?
- CBSE Class IX
- Graphical Representation of Motion
- (i) The kind of person the doctor is (money, possessions)
(ii) The kind of person he wants to be (appearance, ambition)- CBSE Class IX
- The snake and the mirror
- Rewrite the sentences below, changing the verbs in brackets into the passive form.
1. In yesterday’s competition the prizes (give away) by the Principal.
2. In spite of financial difficulties, the labourers (pay) on time.
3. On Republic Day, vehicles (not allow) beyond this point.
4. Second-hand books (buy and sell) on the pavement every Saturday.
5. Elections to the Lok Sabha (hold) every five years.6. Our National Anthem (compose) Rabindranath Tagore.- CBSE Class IX
- My childhood
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.