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Test: Work, Energy & Power - 1 - JEE MCQ


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30 Questions MCQ Test - Test: Work, Energy & Power - 1

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Test: Work, Energy & Power - 1 - Question 1

A particle experiences a variable force in a horizontal x-y plane. Assume distance in meters and force in newton. If the particle moves from point (1, 2) to point (2, 3) in the x-y plane; then kinetic energy changes by

Detailed Solution for Test: Work, Energy & Power - 1 - Question 1

Test: Work, Energy & Power - 1 - Question 2

A stone of mass m tied to a string is being whirled in a vertical circle with a uniform speed. The tension in the string is

Detailed Solution for Test: Work, Energy & Power - 1 - Question 2


At any θ,
T - mgcosθ =

⇒ T = mg cosθ + 
Since v is constant,
⇒ T will be minimum when cos θ is minimum.
⇒ θ = 180° corresponds to Tminimum.

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Test: Work, Energy & Power - 1 - Question 3

A block of mass 'm' (as shown in figure) moving with kinetic energy E compresses a spring through a distance 25 cm when its speed is halved. The value of spring constant of used spring will be nE Nm-1 for n = _____________. (in integers)

Detailed Solution for Test: Work, Energy & Power - 1 - Question 3

Explanation:W = Force x displacement in the direction of forceWork done will be zero because porter is stationary (i.e. displacement is zero)

n = 24

Test: Work, Energy & Power - 1 - Question 4

Two cylindrical vessels of equal cross-sectional area 16 cm2 contain water up to heights 100 cm and 150 cm, respectively. The vessels are interconnected so that the water levels in them become equal. The work done by the force of gravity during the process is: [Take density of water = 103 kg/m3 and g = 10 ms-2]

Detailed Solution for Test: Work, Energy & Power - 1 - Question 4


Test: Work, Energy & Power - 1 - Question 5

A body of mass 1 kg is thrown upwards with a velocity 20m/s. It momentarily comes to rest after attaining a height of 18 m. How much energy is lost due to air friction?
( g=10m/s):

Detailed Solution for Test: Work, Energy & Power - 1 - Question 5

Step 1: Total Initial Energy (Kinetic Energy)

The kinetic energy at the start is given by

:

Substituting the values:

Step 2: Energy at Maximum Height (Potential Energy)

At the maximum height, the body momentarily comes to rest, so the energy is purely potential energy (PE):

PE=mgh

Substituting the values:

PE=(1)(10)(18)=180 J

Step 3: Energy Lost to Air Friction

The energy lost due to air friction is the difference between the initial energy and the energy at the maximum height:

Energy lost = K E − P E

Energy lost = 200 −180 = 20 J

Test: Work, Energy & Power - 1 - Question 6

As per the given figure, two blocks each of mass 250 g are connected to a spring of spring constant 2 Nm-1. If both are given velocity v in opposite directions, then maximum elongation of the spring is:

Detailed Solution for Test: Work, Energy & Power - 1 - Question 6


Using energy conservation,

Test: Work, Energy & Power - 1 - Question 7

A body of mass 0.5 kg travels on a straight line path with velocity v = (3x2 + 4) m/s. The net work done by the force during its displacement from x = 0 to x = 2 m is:

Detailed Solution for Test: Work, Energy & Power - 1 - Question 7

Given,
Mass = 0.5 kg
At x = 0
vi = 3(02) + 4 = 4
At x = 2
vf = 3(2)2 + 4
= 16

Test: Work, Energy & Power - 1 - Question 8

A particle of mass 500 gm is moving in a straight line with velocity v = b x5/2. The work done by the net force during its displacement from x = 0 to x = 4 m is: (Take b = 0.25 m-3/2 s-1)

Detailed Solution for Test: Work, Energy & Power - 1 - Question 8

From the work energy theorem,
Work done by net force = ΔK.E.

w = 16 J

Test: Work, Energy & Power - 1 - Question 9

A body of mass M at rest explodes into three pieces, in the ratio of masses 1 : 1 : 2. Two smaller pieces fly off perpendicular to each other with velocities of 30 ms-1 and 40 ms-1 respectively. The velocity of the third piece will be:

Detailed Solution for Test: Work, Energy & Power - 1 - Question 9

Mass of pieces according to the statement-

According to law of conservation of momentum


so v = 25 ms-1

Test: Work, Energy & Power - 1 - Question 10

A free electron of 2.6 eV energy collides with a H+ ion. This results in the formation of a hydrogen atom in the first excited state and a photon is released. Find the frequency of the emitted photon (h = 6.6 × 10-34 J s).

Detailed Solution for Test: Work, Energy & Power - 1 - Question 10

For every large distance, P.E. = 0
Total energy = 2.6 + 0 = 2.6 eV
Finally, in the first excited state of H atom, total energy = -3.4 eV
Loss in total energy = 2.6 - (-3.4) = 6 eV
It is emitted as photon.

= 1.45 × 109 MHz

Test: Work, Energy & Power - 1 - Question 11

A particle of mass 1 kg is hanging from a spring of force constant 100 Nm-1. The mass is pulled slightly downward and released, so that it executes free simple harmonic motion with time period T. The time when the kinetic energy and potential energy of the system will become equal is The value of x is _______. (in integers)

Detailed Solution for Test: Work, Energy & Power - 1 - Question 11

K E = PE

A2 = 2x2

X = A sinωt

x = 8 sec

Test: Work, Energy & Power - 1 - Question 12

Given below is the plot of a potential energy function U(x) for a system, in which a particle is in one dimensional motion, while a conservative force F(x) acts on it. Suppose that Emech = 8 J, the incorrect statement for this system is:

Detailed Solution for Test: Work, Energy & Power - 1 - Question 12

Emech. = 8 J
(A) At x < x1, U = constant = 8 J
K = Emech - U = 8 - 8 = 0 J
The particle is at rest.

(B) At x = x2, U = 0 ⇒ Emech. = K = 8 J
K.E. is greatest and the particle is moving at fastest speed.

(C) At x = x3, U = 4 J
U + K = 8 J
K = 4 J

Hence the incorrect statement is : at x < x1, K.E is smallest and the particle is moving at the slowest speed.

(D) At x > x4, U = constant = 6 J
K = Emech. - U = 2 J = constant

Test: Work, Energy & Power - 1 - Question 13

A body at rest is moved along a horizontal straight line by a machine delivering a constant power. The distance moved by the body in time 't' is proportional to

Detailed Solution for Test: Work, Energy & Power - 1 - Question 13

P = constant

By integration,

Test: Work, Energy & Power - 1 - Question 14

In a spring gun having spring constant 100 N/m, a small ball 'B' of mass 100 g is put in its barrel (as shown in figure) by compressing the spring through 0.05 m. There should be a box placed at a distance 'd' on the ground so that the ball falls in it. If the ball leaves the gun horizontally at a height of 2 m above the ground, the value of d is _______________ m.
(g = 10 m/s2) (in integers)

Detailed Solution for Test: Work, Energy & Power - 1 - Question 14


Test: Work, Energy & Power - 1 - Question 15

A 60 HP electric motor lifts an elevator having a maximum total load capacity of 2000 kg. If the frictional force on the elevator is 4000 N, the speed of the elevator at full load is close to: (1 HP = 746 W, g = 10 ms-2)

Detailed Solution for Test: Work, Energy & Power - 1 - Question 15

Ftotal = Mg + friction
= 2000 × 10 + 4000
= 20,000 + 4000 = 24,000 N
P = F × v
60 × 746 = 24,000 × v
⇒ v = 1.86 m/s ≈ 1.9 m/s

Test: Work, Energy & Power - 1 - Question 16

In an inelastic collision

Detailed Solution for Test: Work, Energy & Power - 1 - Question 16

Explanation:kinetic energy is transferred to other forms of energy—such as thermal energy, potential energy, and sound—during the collision process. After collision if recovery of kinetic energy is less than % then it is called inelastic collision i.e.. some part of kinetic energy is not recover. So that in an inelastic collision the total kinetic energy after the collision is less than before the collision

Test: Work, Energy & Power - 1 - Question 17

Physically, the notion of potential energy is applicable only to

Detailed Solution for Test: Work, Energy & Power - 1 - Question 17

Explanation:Potential energy is the stored energy of an object. It is the energy by virtue of an object's position relative to other objects. Potential energy is often associated with restoring forces such as a spring or the force of gravity. It is applicable only for conservative forces.

Test: Work, Energy & Power - 1 - Question 18

If a machine is lubricated with oil

Detailed Solution for Test: Work, Energy & Power - 1 - Question 18

When a machine is lubricated with oil friction decreases. Hence the mechanical efficiency of the machine increases.

Test: Work, Energy & Power - 1 - Question 19

A bolt of mass 0.3 kg falls from the ceiling of an elevator moving down with an uniform speed of 7 m/s. It hits the floor of the elevator (length of the elevator = 3 m) and does not rebound. What is the heat produced by the impact?

Detailed Solution for Test: Work, Energy & Power - 1 - Question 19

Explanation:

Whole of the potential energy of bolt converted in to heat energy 

heat produced by the impact = mgh =0.3×9.8×3=8.82J

Test: Work, Energy & Power - 1 - Question 20

The work done by a conservative force

Detailed Solution for Test: Work, Energy & Power - 1 - Question 20

Explanation:A force is said to be conservative if work done by this force is independent of path and is dependent only on end points .

Test: Work, Energy & Power - 1 - Question 21

In which of the following cases is the work done positive?

Detailed Solution for Test: Work, Energy & Power - 1 - Question 21

Explanation:Work done is positive when force applied and displacement are in same direction.

W=F .s =Fscosθ

in this case angle between force and displacement is zero (i.e. cosθ=1) so that work done is positive.

Test: Work, Energy & Power - 1 - Question 22

The total mechanical energy of a system is conserved if the

Detailed Solution for Test: Work, Energy & Power - 1 - Question 22

Explanation:Mechanical energy is the sum of kinetic and potential energy in an object that is used to do work. In other words, it is energy in an object due to its motion or position, or both. In case of conservative forces total mechanical energy remains conserved because potential energy applicable only for conservative forces.

Test: Work, Energy & Power - 1 - Question 23

The casing of a rocket in flight burns up due to friction. At whose expense is the heat energy required for burning obtained?

Detailed Solution for Test: Work, Energy & Power - 1 - Question 23

Explanation:Heat energy required for burning of casing of rocket comes from the rocket itself. As a result of work done against friction the kinetic energy of rocket continuously deceases and this work against friction reappears as heat energy. This results in reduction of the mass of the rocket.

Test: Work, Energy & Power - 1 - Question 24

A tandem (two-person) bicycle team must overcome a force of 165 N to maintain a speed of 9.00 m/s. Find the power required per rider, assuming that each contributes equally.

Detailed Solution for Test: Work, Energy & Power - 1 - Question 24

Explanation:

Total power required to overcome a force of 165 N and to maintain a speed of 9.00 m/s

F = 165N

v = 9m/s

P = Fv = 165 x 9 = 1485 W

if each rider contribute equal power, then power required per rider will be P/2 = 1485/2 = 742.5W

 

Test: Work, Energy & Power - 1 - Question 25

For a ball dropped from a tower of height h the total mechanical energy is

Detailed Solution for Test: Work, Energy & Power - 1 - Question 25

Explanation:mechanical energy = sum of potential and kinetic energiesa falling ball will have both these energies in between topmost and bottomost points of its motion so mechanical energy is the sum of potential and kinetic energies.

Test: Work, Energy & Power - 1 - Question 26

A trolley of mass 200 kg moves with a uniform speed of 36 km/h on a frictionless track. A child of mass 20 kg runs on the trolley from one end to the other (10 m away) with a speed of 4 ms−1 relative to the trolley in a direction opposite to the its motion, and jumps out of the trolley. How much has the trolley moved from the time the child begins to run?

Detailed Solution for Test: Work, Energy & Power - 1 - Question 26

Explanation:
Mass of troly M = 200Kg
mass of child m = 20Kg
speed of trolley v = 36Km/hr=36 x 5/18 = 10m/s
Let v' be the final velocity of the trolley with respect to the ground.
Final velocity of the boy with respect to the ground = v,−4
from conservation of linear momentum

Time taken by the boy to run t = 10/4 = 2.5 sec
Distance moved by the trolley = vt = 10.36 x 2.5 = 25.9 m

Test: Work, Energy & Power - 1 - Question 27

A 50.0-kg marathon runner runs up the stairs to the top of a 443-m-tall Tower. To lift herself to the top in 15.0 minutes, what must be her average power output?

Detailed Solution for Test: Work, Energy & Power - 1 - Question 27

https://edurev.gumlet.io/ApplicationImages/Temp/485050_ee1debcc-c597-4ec1-9241-25680719afee_lg.PNG

Test: Work, Energy & Power - 1 - Question 28

In which of the following cases is the work done positive?

Detailed Solution for Test: Work, Energy & Power - 1 - Question 28

Explanation:When a body is moving on a rough horizontal surface then their will be 2 forces acting on the body   1. Applied force ( in the direction of motion) 2. friction ( opposite to direction of motion)As applied force is in same direction as displacement so work done will be positive.

Test: Work, Energy & Power - 1 - Question 29

The launching mechanism of a toy gun consists of a spring of unknown spring constant. When the spring is compressed 0.120 m, the gun, when fired vertically, is able to launch a 35.0-g projectile to a maximum height of 20.0 m above the position of the projectile before firing. Neglecting all resistive forces, determine the spring constant.

Detailed Solution for Test: Work, Energy & Power - 1 - Question 29

Explanation:

Potential energy of spring converted in to potential energy

Test: Work, Energy & Power - 1 - Question 30

If F is a force and d is the displacement in the direction of force then the work done by the force is given by

Detailed Solution for Test: Work, Energy & Power - 1 - Question 30

Explanation:Work done = force in the direction of displacement multiplied by displacement

W=F.d

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