NEET Work, Energy and Power Previous Year Questions with Solutions

At any intermediate position x along the rough track, the normal reaction on the slab is N = M*g. The instantaneous kinetic friction opposing motion is given by:

• • Because the slab moves slowly at constant speed, external pulling force matches friction at every instant.
• • Infinitesimal work required to advance by dx is dW = f_k(x) dx.

Set up the definite integral from the entry point x = 0 to the exit point x = L:

• • Pull out constant scalar factors (mu_0, M, g, L) outside the integral sign.
• • Evaluate integral of x dx: [x^2 / 2] evaluated from 0 to L.

Substitute upper limit L and lower limit 0 into the algebraic expression:

• • The factor L in the denominator cancels with one factor of L in L^2.
• • Yields the final exact work value: (1/2) * mu_0 * M * g * L.

For an elastic collision between two bodies of equal mass m_A = m_B = m:

• • The incident mass A transfers its entire momentum and kinetic energy to bob B.
• • Bob B begins moving horizontally from the bottom of its arc with speed v_B = 10 m/s.

As bob B swings upward against gravity, kinetic energy converts into gravitational potential energy:

• • Mass m cancels out.
• • h is the vertical rise above the lowest point.

Substitute v_B = 10 m/s and g = 10 m/s^2:

• • Matches Option (3) exactly.

When a car of initial kinetic energy KE is brought to rest by a constant braking force F over distance s, the net work done by friction/brakes equals the loss in kinetic energy.

• • Forces are assumed constant over the stopping distances.

Set up the ratio of the braking forces for cars A and B:

• • 225 / 1500 simplifies by dividing numerator and denominator by 75: 3 / 20 = 0.15.
• • Ratio is exactly 2/3, corresponding to Option (2).

Point P is in the upper quadrant, where the string makes an angle θ above the horizontal. The component of gravity acting radially inward towards the suspension point O is mg sin θ.

• • When the string becomes slack, tension vanishes: T = 0.
• • Therefore, mg sin θ = m v² / l ⟹ v² = g l sin θ.

Taking the bottom-most point as reference potential energy (PE = 0), the height of point P is h = l + l sin θ. Conservation of mechanical energy between the bottom and point P gives:

• • Express gl in terms of v²: gl = v² / sin θ.

Substitute gl into the energy equation:

• • Inverting to find the ratio v / v₀:
• • v / v₀ = ( sin θ / (2 + 3 sin θ) )^(1/2), matching Option (4).

Before the collision, body A of mass m moves with velocity v_1 while body B of identical mass m is at rest. The total initial linear momentum of the system is:

In a completely inelastic one-dimensional collision, coefficient of restitution e = 0. The two bodies coalesce into a single combined mass of 2m moving together with common velocity v_2:

In the absence of external impulsive forces, linear momentum is conserved:

Dividing both sides by m v_2 gives the ratio v_1 : v_2:

The velocity v of the particle is the first time derivative of its position function x(t) = 2t - 1:

Instantaneous power P delivered by a force F to a particle moving with velocity v in the same direction is:

Substituting F = 5 N and v = 2 m/s:

The work done against the restoring spring force F = -kx in stretching a spring of spring constant k by displacement x is stored as elastic potential energy:

Write expressions for initial state x₁ = 2 cm and final state x₂ = 8 cm:

Take the ratio of final to initial potential energy:

Water of mass $\Delta m$ falling from height $h$ carries gravitational potential energy $\Delta U = (\Delta m)gh$. The rate of potential energy delivered (input power) is:

Frictional losses account for $10\%$ of the input energy. The remaining $90\%$ is converted into electrical output power by the turbine:

Multiply input power by the net efficiency factor:

At height $S$ above the ground, the particle is released from rest ($u = 0$). Taking the ground as the reference plane for zero gravitational potential energy:

At an unknown height $h$, potential energy is $U = mgh$ and kinetic energy is $K$. We are given $K = 3U$:

Substitute $h = S/4$ into the expression for kinetic energy $K = \frac{1}{2}mv^2$: