Test: Work, Energy & Power - 2 - JEE MCQ

The person applies ‘F’ constantly in a fixed direction. So, to keep the particle in constant circular motion of radius ‘r’ and uniform speed ‘v’, some other force whose resultant varies in magnitude and direction also act on the particle. Since the resultant of F and other force has to have constant magnitude but varying direction, the magnitude and direction of the other force has to change from point to point on circle.

The tangential acceleration is clearly gsinθ, by simply taking the components of the tension vector.
And the net for on the bob is
F = T – mgcosθ
And it will be equal to centripetal force, so,
T – mgcosθ = Mv²/l

Linear Velocity of rotating object ν=ωr/sinθ
ω= angular velocity=0.1 rad/s
r is the distance of object from centre.
Angle = 45o
r = d/cosθ [where d is the distance from the spot light to wall] = 3/cos45o
Velocity = ν = ωr/sinθ
= [0.1x3/cos45degree]/sin45degree
= 0.1x3/sin45 degreecos45degree
= 0.3/[1√2 x 1 /2]
= 2x0.3
= 0.6m/s
The velocity of the spot P is 0.6 m/s

If a particle is moving with angular velocity=ω
Its angle of rotation is given by ωt 
Now displacement= length of line AB
Position vector of a particle is given by
R =iacosωt + jasinωt
R_o=ai
displacement =R−R_o
=a(cosωt−1)i+asinωj
d=√[(a(cosωt-1))²+(asinω)²]
=a√(2(1-cosωt))=a√(2×2(sinωt/2)²)=2asinωt/2

T= (mv²_max)/r
16 = (16 v²_max)/144
v²_max = 12 m/s

Work done = f.ds
So on perpendicular application for product becomes zero so no work is done
 

Stored elastic potential energy of spring =1/2​kx2 where x is compression or elongation of spring from its natural length. In this position the spring can do work on the block tied to it, which is equal to 1/2​kx2, so both option (a) & b are correct.

Since the motion of the body can be placed in the direction of friction, opposite to the direction of motion and even can not be placed in any motion.
So,  Work done by force of friction can be zero, negative, and can be positive.

If the total energy is E that means that E = U(x) + KE and if KE = 0 then E = U(x)

According to the Work Energy Theorem,
Change in PE = Change in KE
So, mgh = ½ mv² 
Hence C
Forum id- 1775160
For complete circle v = √(5gl)
mgh = ½mv²
So, h = 2.5R

The ball is at rest, it has maximum potential energy. When the ball is released from rest with the spring at its normal (unstretched) length it loses some potential energy and energy of spring increases. Hence, loss in potential energy of the ball is equal to gain in potential energy of spring.
∴mgx=1/2​kx²
∴x=2mg/k​
Also, for x′=x/2​,
kx′=mg i.e. forces are equal thus, the ball will have no acceleration at the position where it has descended through 2x​.
And when ball is at lowermost position, the spring force will be
kx=2mg
Hence, the ball will have an upward acceleration equal to g at its lowermost position.

The particle will land outside the circle because its initial velocity in horizontal direction is in the tangential direction and its point of projection is somewhere at the circumference of the circle.
Also its path will be parabolic because it is a case of projectile motion with velocity in horizontal as well as vertical direction.
 

Similarly you can find that the particle crosses x - axis at (3, 0) with a velocity 7.07 m/s

Velocity on reaching the base of circle is
u=√2gH=√2g²R=√4gR
This is less than √5gR. Hence, the particle cannot complete the circle. But this velocity is more than √2gR so the particle will break off the loop at a height above the centre before reaching the top.
 

For complete circle v = √(5gl)
mgh = ½mv²
So, h = 2.5R

Since friction is present in the scenario, so, only WET can be applied.

The question is incomplete and doesn’t make sense by itself. It should be changed to:
“A spring block system is placed on a rough horizontal floor. The block is pulled towards right to give spring an elongation less than 2μmg/K but more than μmg/K and released.
The correct statement is:”
Also add this image

Initially the spring block system will have only one form of energy and that will be spring potential energy stored in it. After the system is released from this position it starts to oscillate to and fro with friction acting on it. When the system comes to rest then the total stored potential energy will be converted into the work done by the frictional force on the block.