Question:
An object of mass $10\, kg$ falls from rest through a vertical distance of $10\,m$ and acquires a velocity of $10\,m/s$ .The work done by the push of air on the object is $(g = 10 \, m/s^2)$
An object of mass $10\, kg$ falls from rest through a vertical distance of $10\,m$ and acquires a velocity of $10\,m/s$ .The work done by the push of air on the object is $(g = 10 \, m/s^2)$
Updated On: Jun 20, 2022
- 500 J
- - 500 J
- 250 J
- - 250 J
Hide Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Given, mass of object $=10\, kg$
Initial velocity (rest), $u =0$
Final velocity, $v =10 m / s$
$g =10\, m / s ^{2}$
Work done by air $=-($ Work done by the object to cover the vertical distance)
$=- F \times s =- mgh [as, v ^{2}- u ^{2}=2 gh \Rightarrow gh =\frac{ v ^{2}- u ^{2}}{2}]$
$=-10 \times \frac{\left(10^{2}-0^{2}\right)}{2}=-\frac{10 \times 10 \times 10}{2}=-500\, J$
Initial velocity (rest), $u =0$
Final velocity, $v =10 m / s$
$g =10\, m / s ^{2}$
Work done by air $=-($ Work done by the object to cover the vertical distance)
$=- F \times s =- mgh [as, v ^{2}- u ^{2}=2 gh \Rightarrow gh =\frac{ v ^{2}- u ^{2}}{2}]$
$=-10 \times \frac{\left(10^{2}-0^{2}\right)}{2}=-\frac{10 \times 10 \times 10}{2}=-500\, J$
Was this answer helpful?
0
0
Top Questions on Work-energy theorem
- A block of mass \(5 kg\) moves along the \(x-\)direction subject to the force \(F = (−20x + 10) N,\) with the value of \(x \) in metre. At time \(t = 0 s,\) it is at rest at position \(x = 1 m\). The position and momentum of the block at \(t = (\pi/4)\) s are
- JEE Advanced - 2024
- Physics
- Work-energy theorem
- The energy associated with electric field is E and with magnetic field is B for an electromagnetic wave in free space is given by( \(∈_0\)-permittivity of free space \(μ_0\)-permeability of free space)
- JEE Main - 2023
- Physics
- Work-energy theorem
- An electric bulb is rated as 200 W. What will be the peak magnetic field at 4 m distance produced by the radiations coming from this bulb? Consider this bulb as a point source with 3.5% efficiency.
- JEE Main - 2022
- Physics
- Work-energy theorem
- The light of two different frequencies whose photons have energies 3.8 eV and 1.4 eV respectively, illuminate a metallic surface whose work function is 0.6 eV successively. The ratio of maximum speeds of emitted electrons for the two frequencies respectively will be
- JEE Main - 2022
- Physics
- Work-energy theorem
- A 100 g of iron nail is hit by a \(1.5\) \(kg\) hammer striking at a velocity of \(60 \;ms^{–1}\). What will be the rise in the temperature of the nail if one fourth of energy of the hammer goes into heating the nail?
[Specific heat capacity of iron = \(0.42\; Jg^{–1} °C^{–1}\)]- JEE Main - 2022
- Physics
- Work-energy theorem
View More Questions
Questions Asked in UPSEE exam
- A carnot engine takes 300 calories of heat at 500 K and rejects 150 calories of heat to the sink. The temperature of the sink is
- UPSEE - 2019
- Heat Transfer
- Which of the following is not permissible arrangement of electrons in an atom.
- UPSEE - 2019
- Quantum Mechanical Model of Atom
- If det $\begin{bmatrix}1&1&2\\ 2&4&9\\ t&t^{2}&1+t^{3}\end{bmatrix} = 0 $ , then the values of t are
- UPSEE - 2019
- Properties of Determinants
- The imaginary part of $\left( \frac{1}{2} + \frac{1}{2}i\right)^{10} $ is
- UPSEE - 2019
- Complex Numbers and Quadratic Equations
- Let S be the set of all right angled triangles with integer sides forming consecutive terms of an arithmetic progression. The number of triangles in S with perimeter less than 30 is
- UPSEE - 2019
- Arithmetic Progression
View More Questions
Concepts Used:
Work-Energy Theorem
The work and kinetic energy principle (also known as the work-energy theorem) asserts that the work done by all forces acting on a particle equals the change in the particle's kinetic energy. By defining the work of the torque and rotational kinetic energy, this definition can be extended to rigid bodies.
The change in kinetic energy KE is equal to the work W done by the net force on a particle is given by,
W = ΔKE = ½ mv2f − ½ mv2i
Where,
vi → Speeds of the particle before the application of force
vf → Speeds of the particle after the application of force
m → Particle’s mass
Note: Energy and Momentum are related by, E = p2 / 2m.