HC Verma Test: Work, Energy & Power - NEET MCQ

∴ Spring is compressed by 2 cm at A and will be stretched by 2 cm at B.
By work-energy theorem,
W_mg + W_s + W_N + W_f = K_f − K_i
0.16 + 0 + 0 + W_f = 0 - 10
W_f = -10.16 J

Given: D = 2R = 42m, R = 21m
h = R(1 − cosθ)
From the 3rd equation of motion,

v² = u² + 2gh
v² = 2gR(1−cosθ)  ...(1)

From the free body diagram:

When contact lost, N = 0, equation (2) becomes:

mgcosθ = 2mg(1−cosθ)
cosθ = 2−2cosθ
3cosθ = 2
cosθ = 2/3​

 2R − h = 2R − R(1−cosθ)
2R − h = 2R − R(1−2/3​)
5R/3​ = 5/3 ​× 21 = 35m

• Since the given chain is uniform we can find the mass of the hanging part of the chain if we define some quantity lambda as mass per unit length and then multiplying it with the length of the hanging part of the chain.
• Also, while applying the energy conservation equation for extended bodies (like chain, etc) we need to consider the energies (both kinetic and potential) of the center of mass only. We will consider the dotted line as a reference line while applying energy equation.

• As the particle is projected with some velocity, therefore, its initial kinetic energy will not be zero.
• As it moves downward under gravity then its velocity increases with time K.E. ∝ v² ∝ t² (As υ ∝ t) 
• So the graph between kinetic energy and time will be parabolic in nature. 

3/4 th energy is lost i.e. 1/4 th kinetic energy is left.

Hence its velocity  becomes v₀/2 under a retardation of μg in time t₀.

Let the velocity at point B be v.

By conservation of energy, mgh = (1/2 ) mv² ⇒ v = √(2gh​)
Now deceleration due to friction is a = μg = 0.2 g
So stopping distance, s = v²/(2a) ​= (2gh) / (0.4g) = 5 h = 5 × 1 = 5m

N sinθ = kx
N cosθ = mg
We get the value of x, 
x = (mg tanθ) / k

The slope of the energy distance graph represents force.

∵ At points B, C slope is zero,
Therefore, force acting is also zero

We have two blocks with upper block in equilibrium. The upper block compresses the spring by 1 cm.

Therefore force by spring is equal to the weight of 2 kg block.
kx = 2g ⇒ k × 1 = 2g ⇒ k = (2g)/1 = 20 N/cm = 2000 N/m

Now let the total compression after exerting force be y. When the force is removed the upper block goes up lifting the lower block upward.
At this time the normal reaction force acting on the body becomes zero.
This means spring force will be equal to the sum of the weights of upper block and lower block.

Ky = 2g + 3g
Replacing k by 2000N/m and g by 10m/s²
2000 y = 2 × 10 + 3 × 10 = 50
⇒ y = 50/2000
⇒ y = 0.025 m = 0.025 x 100
⇒ y = 2.5 cm

Velocity at point C is v

Let the Normal Reaction of Wedge on the Block be 'N'.

We know,
Acceleration of Block , a = 10 m/s²
Initial velocity of block , u = 0 m/s

Using Newton's Equation of motion,
T=√3s
S = ut + (1/2) at² = 0 + (1/2) x 10 x (√3)² = 15

Work done by the Normal Force,W_N = N x S x cos(π/2- θ)
= N x 15 x sin(θ) = 10 x 15 = 150N

Hence, Work Done by the Normal Force in the ground Frame on the block is 150 N.

Work done on m=change in kinetic energy
⇒ work done by friction and PE energy in spring= Δk.E
⇒ μn(x)+1/2kx²=1/2mv²
⇒ 0.25 x 10 x 4 + 1/2 x 275/100 x 16 = 1/2 x (1) v²
⇒10+22=1/2v²
⇒ 32x2=v²
⇒ 64=v²
⇒ v=8m/s

The block A is given 10m/s velocity
Initial momentum of the system = 3 x 10 = 30
As no external force is applied the linear momentum of the system: 30=6xv
v=5 m/s
The final mass will be taken 6 kg as it is given that finally, all the blocks are moving together as a system.

By the conversation of energy
(1/2) x 3 x 10² + frictional work done = (1/2) x 6 x 5²
⇒ Frictional work done = -75J

work done by acc of sphere = m × 2gR sin∅
work done by gravity = - m × gR(1 - cos ∅)
∆KE = 0
using work-energy threorm
2 mgR sin ∅ - mgR(1 - cos ∅) = 0
2 mgR sin ∅ = mgR(1 - cos ∅)
2 sin ∅ = (1 - cos ∅)
∅ = 2 tan^-12

The work done when moving an object from B to A against friction can be understood as follows:

• Friction work when sliding down from A to B is -mgh. This represents the energy lost due to friction.
• To move the object slowly from B to A, you need to overcome this friction and also provide energy equivalent to the gravitational potential energy difference, which is mgh.
• Therefore, the total work done when moving the object from B to A is the sum of both these energies: -(-mgh) + mgh = 2mgh.

Thus, the work done in moving the object from B to A is 2mgh.

Given,
Pressure, P = 20000 Nm^-2
Volume per second = 1cc
Power of heart = Pressure x volume per second
= 20000 Nm^-2 x 10^-6 = 0.02 w

According to the work-energy theorem, the total work done by the force of gravity and the force of air resistance on the object is equal to the change in kinetic energy.
Work done by all forces = Change in KE

The sum of work done by force of friction and spring force is equal to change in kinetic energy of the block.
Suppose the block comes to rest at the point E, i.e., let DE = x. The kinetic energy of the block is spent in overcoming friction and compressing the spring through a distance DE = x.
Kinetic energy of the block
....(i)
As the part AB of the track is frictionless, work done in moving from A to B is zero. 
Let normal reaction of the block = mg.
C oefficient of friction = µ
Force due to friction along the track
BC = µ mg = 0.2 x 0.5 x 10 = 1N
Distance through which the block moves against the frictional force = 2.14 + x m

Work done by block against friction before it comes to rest = µ mg (2.14+ x); = (2.14 + x) J
Let the spring constant = k
∴ Work done by the block in compressing the spring through distance X
...(iii)
Adding (ii) and (iii) and equating it to (i), we get
2.14 + x + x² = 2.25; or x² +  x - 0.11 = 0
or 100x² + 100x - 11 = 0
or (10x + 11)(10x - 1) = 0

Restoring force of the spring
= kx = 2 x 0.1 = 0.2N … (iv)
Static frictional force of the block
µ_static mg= 0.22 x 0.5 x 10 = 1.1N … (v)
From (iv) and (v) it is clear that the static frictional force is greater than the restoring force of the spring. Therefore, the block will not move in the backward direction. Hence the total distance through which the block moves before it comes to rest completely is
2.00 + 2.14 + 0.10 = 4.24 m

Apply newton’s second law of motion along the direction of incline to find the applied force. As acceleration is constant and initial velocity is zero, the final velocity at time t is v=at along incline and power is P = F× v.
Resolving forces parallel to incline
F − mg sin θ= ma ; ⇒ F = mg sin θ + ma
= 2 x 9.8x sin30^o + 2x1= 11.8N
The velocity after t = 4 s is v = u+at = 0 + 1x 4= 4m / s
Power delivered by force at t = 4 s
= Force x Velocity = 11.8 x 4 = 47.2W
The displacement during t = 4 s is given by the formula
v² = u² + 2as; v² = 0 + 2 × 1 × S
∴ S = 8m
Work done in t = 4s is W = Force × distance = 11.8 x 8 = 94.4 J
∴  average power delivered

As the spring expands the potential energy stored in the spring is converted in the kinetic energy of ball. The horizontal distance moved by the ball will depend on time taken by ball to fall vertical height h

 When the spring is released its elastic potential energy is converted into kinectic energy 

 As vertical component of velocity of the ball is zero, the time taken by the ball to reach the ground is

Therefore, the horizontal distance traveled by the ball in this time is

As the blocks A and C fall vertically downwards, the loss in its potential energy is equal to gain in kinetic energy of blocks A, B and C.
Assume
m_A = m_B 200kg and m_c = 100kg
At the instant, when the bar is as shown in the Figure

where dx/dt = velocity of B and dy/dt = velocity of A
Applying the law of conservation of energy, loss of potential energy of A, if it is going down when the rod is vertical to the position as shown in the Fig.

= m_Ag (10 − 8) = 2 x 200 x 9.8
C moves down 6 m since B moves 6 m along x-axis.
Total loss of potential energy
= 200 x 9.8 x 2 − 100 x 9.8 x 6 = 100 x 9.8 x 10 = 9800 J.
This must be equal to kinetic energy gained Kinetic energy gained


∴ Veloctiy of B at the required moment is = 6.9ms^-1