Work, Energy & Power - 1 - Free MCQ Practice Test with solutions, NEET

The change in kinetic energy of a particle moving in a horizontal plane can be determined using the work-energy theorem. The theorem states that the work done by a force on a particle equals the change in its kinetic energy. We Know that, 

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 T_min.

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

n = 24

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

Given Data:

Mass of each block: m = 250 g = 0.25 kg,

Spring constant: k = 2 N / m,

Initial velocities of blocks: v (in opposite directions).

When the two blocks move with velocity v in opposite directions, the system's kinetic energy is converted into potential energy stored in the spring at the point of maximum elongation.

The total kinetic energy of the two blocks is:

The potential energy stored in the spring at maximum elongation (x) is:

At maximum elongation, all the kinetic energy of the system is converted into potential energy:

Solution:

• Mass: 0.5 kg
• Initial velocity at x = 0:
  • Formula: v = 3x² + 4
  • Substitute: v = 3(0)² + 4 = 4 m/s
• Final velocity at x = 2:
  • Substitute: v = 3(2)² + 4 = 16 m/s
• Calculate the change in kinetic energy:
  • Initial kinetic energy: (1/2) × 0.5 × 4² = 4 J
  • Final kinetic energy: (1/2) × 0.5 × 16² = 64 J
• Net work done by the force:
  • Work done = Change in kinetic energy = 64 J - 4 J = 60 J

Therefore, the net work done by the force is 60 J.

Solution:

The work-energy theorem states that the work done by a net force is equal to the change in kinetic energy.

Given the velocity equation: v = b x^5/2, where b = 0.25 m^-3/2 s^-1.

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

w = 16 J

The problem involves a body moved by a machine with constant power.

P = constant

By integration,

F_total = 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

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

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.

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

Given:

• Angle of projection, θ = 30°
• Initial kinetic energy = 40 J

Concept: At the topmost point of a projectile’s motion, the vertical component of velocity becomes zero. So the kinetic energy is only due to the horizontal component.

Let initial speed be u, mass be m.
Total kinetic energy:
KE = (1/2)mu² = 40 J ⇒ mu² = 80

At the top: only horizontal velocity remains:
u_x = u cosθ ⇒ KE_top = (1/2)mu²cos²θ

cos 30° = √3/2 ⇒ cos²30° = 3/4
KE_top = (1/2 × 80 × 3/4) = 30 J

Answer: (d) 30 J

Option C is correct.

The frictional force equals μmg, so the relative deceleration of the body due to friction is a = μg.

With μ = 0.2 and g = 10 m/s², a = 0.2 × 10 = 2 m/s².

The initial relative speed is u = 4 m/s and the final relative speed is v = 0.

Use the kinematic relation v² = u² - 2as.

Substituting values: 0 = (4)² - 2 × 2 × s.

So s = 16 / 4 = 4 m.

Therefore the body slides a distance of 4 m relative to the platform; option C is correct.

F = 10× 12 = 120 N
F = kx = 2400 x
∴ x = 1/20
Energy stored in the spring E = 1/2 Kx²
= 1/2 × 2400 × 1/400 = 3J

​

Let F be the force applied as shown in the Figure.
If F moves through x, the work done will be F× x.
This is used to work against friction mm1g and store energy in the spring.

Work done against friction = μm₁gx;
Energy stored = 1/2 kx², where k is the spring constant.

When the mass m₂moves, the tension in the spring balances the force of friction at m₂.
∴ kx = μm₂g
∴ combining equations (i) and (ii),

Given: ​

The power developed by the gun is given by formula –

Power = 12250 watt

The forces are shown in Figure.

Net force to move the engine up the slope. 
F = μN + mg sin θ
= mg (μ cos θ + sin θ)
If the engine has to apply an upward force equal to F, power of engine, P = Fv
where v is the velocity equal to 100 m/hr.
Work done by engine, W = Pt  = Fvt
Efficiency of engine,
Energy used by engine 
m = 100000 kg, μ = 0.1, θ = 5°, v = 100 m/hr, t = 1 hr 
η = 4/100 = 0.04
Energy used by engine 

As 1 kg coal yields 50000 J, we have the amount of coal burnt up

The energy possessed by an object due to its height is called gravitational potential energy. It can be calculated using the formula:

• Potential Energy (PE) = mass (m) × gravitational acceleration (g) × height (h)

In this case:

• mass (m) = 10 kg
• gravitational acceleration (g) = 10 m/s²
• height (h) = 20 m

Now, substituting the values into the formula:

• PE = 10 kg × 10 m/s² × 20 m
• PE = 2000 kg·m²/s²

Since 1 kg·m²/s² is equal to 1 Joule, we can convert:

• PE = 2000 J

Therefore, the energy possessed by the object at that height is 2000 J, which is equivalent to 2 kJ.