Work, Power And Energy MCQ Level - 1 - Physics MCQ

The correct answer is: the velocity of the ring when it reaches the ground will be  

We are tasked to determine the mass mm of a body dropped from a height when it loses energy U and acquires a velocity v.
Kinetic Energy and Energy Loss Relationship:
The kinetic energy of the body when it acquires velocity v is:

This kinetic energy comes from the loss of potential energy, which is equal to U.
Equating Energy Loss and Kinetic Energy:

Solve for m:
Rearrange the equation to find the mass m:

We need to determine how the distance moved by the body varies with time t under the condition of constant power.
Relation Between Power and Velocity:
Power delivered by the machine is constant, and power is defined as:
P=F⋅v
Here, 
F=ma and a is acceleration.
Since P is constant, and v = dx/dt:
P=m⋅a⋅v
Using we can integrate further.

Constant Power and Kinetic Energy Relation:
Power is also the rate of change of kinetic energy:

This gives:

Rearranging and integrating for v:
v∝t ^1/2
Distance Moved in Time t:
The velocity v = dx/dt, so:
dx∝t ^1/2 dt
Integrating with respect to t:
x∝t ^3/2

We are tasked to find the distance traveled by a self-propelled vehicle of mass m when its velocity changes from v₁  to v₂, given that it has constant power P.
Power and Acceleration Relationship:
Power is given by:
P=F⋅v
where F is the force acting on the vehicle. Using F=ma:
P=ma⋅v
Acceleration can be expressed as:

Kinematic Relation:
From kinematics, we know:

and velocity v is related to displacement x by:
vdv=adx
Substituting 

Rearranging:

Integrating for Displacement:
To find the total distance traveled x as the velocity changes from v₁ to v₂

The integral of v² is:

Substituting the limits:

We are tasked to find the maximum positive work that the applied force F can do on a block attached to a spring.
Work Done by Force F:
Work done by 
F is given by:
W=F⋅x
Here, x is the displacement of the block.
Force Balancing at Maximum Extension:
At the maximum extension of the spring, the restoring force of the spring balances the applied force:
F=k⋅x
where k is the spring constant.
Rearrange to find x:

Potential Energy Stored in the Spring:
At maximum extension, the spring stores potential energy, which is given by:

Substituting 


Maximum Work Done by F:
The work done by the force F is equal to the energy stored in the spring at maximum extension, since the surface is smooth and no energy is lost:
W _max =U _spring
Substituting the value of U_spring :

We are tasked to find the speed of the block as it passes through the mean position after being displaced through a distance x and released.
Energy Conservation Principle:
The system is conservative, and the total mechanical energy remains constant. At the maximum displacement 
x, all the energy is stored as potential energy in the springs. As the block passes through the mean position, all the potential energy is converted into kinetic energy.
Potential Energy Stored in the Springs:
The two springs are attached in parallel, so the effective spring constant k eff is:
k eff =k₁ +k₂
​The potential energy stored in the springs at displacement x is:

Kinetic Energy at the Mean Position:
At the mean position, the potential energy becomes zero, and all the energy is converted into kinetic energy:

Equating Potential and Kinetic Energy:
Using energy conservation:

Simplify for v² :

Simplify for v ² :

Speed of the Block:
Taking the square root:

We are tasked to determine the distance the block travels when the net work done on it is zero.
Forces Acting on the Block:
The force acting on the block is given as:
F=a−bx²where a>μmg, and μmg is the frictional force acting against the motion.
Work Done by the Forces:
The work done by the applied force F:

The work done against friction W friction:
W friction =−μmg⋅x
Condition for Zero Net Work:
For the net work to be zero:
W _applied+W _friction =0
Substitute the expressions:

Simplify:

Divide through by x (for ≠ 0):

Solve for x²:

Distance Traveled x:
Taking the square root:

We are tasked to find the work done by the man in placing a chain (of mass m and length l) on the table when half of the chain is in the vertical position.
Initial Potential Energy of the Chain:
When the chain is completely vertical, the center of mass of the chain is located at l/2 from the bottom. The initial potential energy is:

Final Potential Energy of Half the Chain on the Table:
When half the chain is on the table, only the other half of the chain contributes to the potential energy. For this half, the center of mass is located at l/4 from the bottom. The final potential energy is:

Work Done by the Man:
The work done by the man is equal to the change in potential energy:
W=U initial −U final
Substituting the values:

Simplify:

We are tasked to determine the potential energy at the highest point of the motion when a body is projected with kinetic energy k at an angle ϕ with the vertical.
Given Information:
Kinetic energy at the point of projection: 
The body is projected at an angle ϕ with the vertical.
Horizontal and Vertical Components of Velocity:

• The vertical component of the velocity: v y =vcosϕ
• The horizontal component of the velocity: v x=vsinϕ

Energy at the Highest Point: At the highest point:
The vertical velocity becomes zero.
The horizontal velocity v_x =vsinϕ remains constant.
Kinetic energy due to the horizontal velocity is:


Conservation of Energy: The total energy (kinetic + potential) at any point remains constant. At the highest point:
Total energy=K _horizontal +U
Since the total energy at the projection point is k, and the horizontal kinetic energy at the highest point is ksin ²ϕ, the potential energy U is:
U=k−K horizontal
Substituting K _horizontal =ksin ² ϕ:
U=k−ksin ² ϕ
U=kcos ² ϕ

We are tasked to find the minimum mass of the ball A (denoted as m _A) so that the block B of mass M just leaves contact with the ground.
Key Concept: For block B to leave the ground, the tension in the spring at the maximum elongation must exceed the weight of block B. This means:
T>Mg
The tension T in the spring is proportional to the displacement x, which depends on the mass of A.
Spring and Energy Analysis: When ball A is released from rest, it moves downward, causing the spring to stretch. The potential energy of A is converted into the elastic potential energy of the spring.
The total elastic potential energy stored in the spring at maximum elongation is:

where x is the maximum displacement and k is the spring constant.
The initial gravitational potential energy of A is:
U_gravity​ =m Agx
From energy conservation:

Solving for kx:
kx=2m Ag

Condition for Block 
B to Leave the Ground: The tension in the spring T is given by:
T=kx
For B to leave the ground:
Substituting T=kx:
2m _Ag>Mg
Cancel g from both sides:
2m A >M