Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Work Power and Energy question

2023 · 31 Jan · Shift 1 · Q65
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Physics
  4. /Work Power and Energy
  5. /2023 · 31 Jan · Shift 1 · Q65

Work Power and Energy question

2023 · 31 Jan · Shift 1 · Q65

JEE MainPhysicsWork Power and EnergyNumerical+4 / −1
A lift of mass M=500 kg\mathrm{M}=500 \mathrm{~kg}M=500 kg is descending with speed of 2 ms−12 \mathrm{~ms}^{-1}2 ms−1. Its supporting cable begins to slip thus allowing it to fall with a constant acceleration of 2 ms−22 \mathrm{~ms}^{-2}2 ms−2. The kinetic energy of the lift at the end of fall through to a distance of 6 m6 \mathrm{~m}6 m will be ‾\underline{\hspace{2cm}}​kJ\mathrm{kJ}kJ.
Numerical answer
View written solutionFree

Correct answer: 7

  1. Given data
  • Mass of lift: M=500 kgM = 500\,\text{kg}M=500kg
  • Initial speed downward: u=2 m s−1u = 2\,\text{m s}^{-1}u=2m s−1
  • Constant downward acceleration: a=2 m s−2a = 2\,\text{m s}^{-2}a=2m s−2
  • Distance fallen: s=6 ms = 6\,\text{m}s=6m

We need the kinetic energy after falling 6 m.

  1. Find final speed using kinematics

Using

v2=u2+2asv^2 = u^2 + 2asv2=u2+2as

Substitute the values:

v2=(2)2+2(2)(6)v^2 = (2)^2 + 2(2)(6)v2=(2)2+2(2)(6) v2=4+24=28v^2 = 4 + 24 = 28v2=4+24=28

So,

v=28 m s−1v = \sqrt{28}\,\text{m s}^{-1}v=28​m s−1
  1. Compute final kinetic energy

Kinetic energy is

K=12Mv2K = \frac{1}{2}Mv^2K=21​Mv2

Substitute M=500M=500M=500 kg and v2=28v^2=28v2=28:

K=12(500)(28)K = \frac{1}{2}(500)(28)K=21​(500)(28) K=250×28=7000 JK = 250 \times 28 = 7000\,\text{J}K=250×28=7000J

Convert to kJ:

7000 J=7 kJ7000\,\text{J} = 7\,\text{kJ}7000J=7kJ
  1. Final answer

The kinetic energy of the lift after falling 6 m6\,\text{m}6m is

7 kJ\boxed{7\,\text{kJ}}7kJ​

So the required integer answer is 7\boxed{7}7​.

  1. Comparison with stored correct answer

Stored correct answer = 777

Our derived answer = 777

They match.

PreviousNext

More from Work Power and Energy

  • A particle experiences a variable force F=(4xi+3y2j​) in a horizontal x-y plane. Assume distance in meters and force is newton. If the particle moves from point (1, 2) to point (2,…2022 · MCQ
  • A ball of mass 100 g is dropped from a height h = 10 cm on a platform fixed at the top of a vertical spring (as shown in figure). The ball stays on the platform and the platform is depressed by a distance 2h​. The spring constant… Includes diagram2022 · Numerical
  • A body of mass 0.5 kg travels on straight line path with velocity v=(3x2+4)m/s. The net workdone by the force during its displacement from x=0 to x=2 m is :2022 · MCQ
  • A bag of sand of mass 9.8 kg is suspended by a rope. A bullet of 200 g travelling with speed 10 ms − 1 gets embedded in it, then loss of kinetic energy will be :2022 · MCQ
  • A uniform chain of 6 m length is placed on a table such that a part of its length is hanging over the edge of the table. The system is at rest. The co-efficient of static friction between the chain and the surface of the table is 0.5, the…2022 · Numerical
  • A 0.5 kg block moving at a speed of 12 ms − 1 compresses a spring through a distance 30 cm when its speed is halved. The spring constant of the spring will be ​ Nm − 1.2022 · Numerical
  • As per the given figure, two blocks each of mass 250 g are connected to a spring of spring constant 2Nm−1. If both are given velocity v in opposite directions, then maximum elongation of the spring is : Includes diagram2022 · MCQ
  • Arrange the four graphs in descending order of total work done; where W1, W2, W3 and W4 are the work done corresponding to figure a, b, c and d respectively. Includes diagram2022 · MCQ