A simple pendulum is released from A as shown. If m and l represent the mass of the bob and length of the pendulum, the gain in kinetic energy at B is
From Principal of conservation of Energy in absence of non-conservative forces,
Ui+ Ki= Uf+ Kf
⇒ Ui- Uf= Kf - Ki
⇒ Loss in Potential Energy = Gain in Kinetic Energy
⇒ Gain in Kinetic Energy = m.g. l .cos 30° = √3mgl/2
If potential energy between electron and proton at a distance r is given by U =-(ke²/3r³), then force acting is
Force F = - dU/dr = d ((ke²/3r³))/dr = ke²/r^4
The potential energy of a particle varies with distance x as shown in the graph :
The force acting on the particle is zero at
For Equilibrium dU/dx =0 so, Points of maxima and minima in U-x graph are point of equilibrium.
So, Points B and C are equilibrium positions
A particle moves with a velocity \((5\hat{i} – 3\hat{j} + 6\hat{k})\text{ m s}^{-1}\) horizontally under the action of constant force \((10\hat{i} + 10\hat{j} + 20\hat{k})\text{ N}\). The instantaneous power supplied to the particle is :
Instantaneous power is the dot product of force and velocity: \(P = \vec{F} . \vec{v} = (10)(5) + (10)(-3) + (20)(6) = 50 - 30 + 120 = 140\text{ W}\).
If the kinetic energy of a body increases by 800%, its momentum increases by
The momentum \(P\) is related to kinetic energy \(K\) by \(P = \sqrt{2mK}\). An 800% increase means the new kinetic energy is \(K' = 9K\). Thus, the new momentum is \(P' = \sqrt{9} P = 3P\), representing an increase of 200%.
A particle with constant total energy \( E \) moves in one dimension in a region where the potential energy is represented by \( U(x) \). The speed of the particle is zero where:
Total mechanical energy of a particle is the sum of its kinetic energy and potential energy: \( E = K + U(x) \). When the speed of the particle is zero, its kinetic energy \( K = 0 \), which gives \( E = U(x) \).
Assertion: Work done by a force depends on frame of reference.
Reason: Force and displacement both depend on frame of reference.
Work done depends on the frame of reference because displacement depends on the frame of reference. However, real forces do not depend on the frame of reference. Hence, the Assertion is true but the Reason is false.
Two masses \( 4m \) and \( 9m \) move with equal kinetic energy. The ratio of the magnitude of their momenta is:
Since kinetic energy \( K \) is the same for both masses, the momentum is proportional to the square root of the mass, \( p = \sqrt{2mK} \). Thus, the ratio of their momenta is \( \frac{p_1}{p_2} = \sqrt{\frac{4m}{9m}} = \frac{2}{3} \).
A force \( \vec{F} = (3x\hat{i} + 4\hat{j})\text{ Newton} \) (where \( x \) is in metres) acts on a particle which moves from a position \( (2\text{ m}, 3\text{ m}) \) to \( (3\text{ m}, 0\text{ m}) \). Then the work done is:
The work done by the force is \( W = \int_{2}^{3} 3x \, dx + \int_{3}^{0} 4 \, dy = \left[ \frac{3x^2}{2} \right]_2^3 + [4y]_3^0 = 7.5 - 12 = -4.5\text{ J} \).
A car of mass \(1000\text{ kg}\) moving at \(20\text{ m/s}\) is brought to rest in \(5\text{ s}\). The average power dissipated due to braking is
The initial kinetic energy of the car is \(K_i = \frac{1}{2}mv^2 = \frac{1}{2}(1000)(20)^2 = 2 \times 10^5\text{ J}\) and final kinetic energy is zero. Average power dissipated is the work done divided by time: \(P = \frac{\Delta K}{t} = \frac{2 \times 10^5\text{ J}}{5\text{ s}} = 40\text{ kW}\).