AIIMS Physics Work Energy And Power Class 11 Questions

A body of mass 5×10³ kg moving with speed 2 m/s collides with a body of mass 15×10³ kg in elastically and sticks to it. Then loss in K.E. of the system will be:

Consider the given expression. F = (100 - 0.5 × 10⁵ t) N Given that, F = 0 (100 - 0.5 × 10⁵ t) = 0 t = 100 / (0.5 × 10⁵) = 2 × 10⁻³ sec Calculate the impulse as, I = ∫ F dt = ∫ (100 - 0.5 × 10⁵ t) dt = [100t - (10⁵ t² / 2)] = [100(2 × 10⁻³) - (10⁵ (2 × 10⁻³)² / 2)] = 0.1 Ns

Use energy method. 1/2 Kx² + 1/2 mv² + 1/2 Iω² = C 1/2 K(2x) dx/dt + 1/2 m(2v) dv/dt + 1/2 2v/r² dv/dt = 0 Kvx + m/4 va + m/2 va =⁰⁻Kx = 3ma/4 Simplify further.

Initially spring in its natural length now a block at mass 0.25 kg is released then find out maximum force by system on floor?

Initially spring in its natural length now a block at mass 0.25 kg is released than find out maximum force by system on floor?

A particle moves from A to B to C as shown in the diagram. Calculate the work done if the force is conservative.

Kinetic energy can be increased or decreased without applying any external force, as internal force can do work. Example: explosion of bomb.

Force exerted by concrete floor is more because change in momentum is fast.

The work done is, W = 1/2 × stress × strain W = 1/2 × E × (strain)² Since the elasticity of steel is more than the copper. Therefore, more work done has to be done in order to stretch the steel.

The potential energy is converted into kinetic energy during sliding whereas in rolling; the some part of potential energy is converted kinetic energy, thereby decreases its kinetic energy of translation. Thus, velocity acquired by the body is less.

A 10 m long iron chain of linear mass density 0.8 kg/m³ is hanging freely from a rigid support. If g = 10 m/s², then the

The ratio of potential and kinetic energies, PE / KE = 2 / 3. mgh / (E - mgh) = 2 / 3 (∵ E = KE + PE) E = 5mgh / 2 ...(1) When the velocity doubled, then the energy becomes 4E. Thus, PE / KE = mgh / (4E - mgh) = mgh / (10mgh - mgh) (from(1)) = 1 / 9

When the load on a wire is increasing slowly from 2 kg to 4 kg, the elongation increases from 0.6 mm to 1 mm. The work done during this extension of the wire is (g = 10 m/s²)

A 40g mass is released from rest while situated at a height 5 cm on the curved track. The minimum deformation in the spring is nearly equal to (take g = 10 m/s²)

The force on a particle as the function of displacement x (in x-direction) is given by F = 10 + 0.5x. The work done corresponding to displacement of particle from x = 0 to x = 2 unit is

In case, S = 0 and so W = 0

A force $\vec{F} = -k\left(y\hat{i} + x\hat{j}\right)$, where $k$ is a positive constant, acts on a particle moving in the $xy$-plane. Starting from the origin, the particle is taken along the positive $x$-axis to the point $(a,0)$ and then parallel to the $y$-axis to the point $(a,a)$. The total work done by the force on the particle is

A conductor lies along the x-axis at $-1.5 \le Z \le 1.5\text{ m}$ carries a fixed current of $10.0\text{ A}$ in $-a_z$ direction as shown in the figure for the field $B = 3 \times 10^{-4} e^{-0.2x} a_y\text{ T}$, the total power required to move the conductor at constant speed to $x = 2.0\text{ m}, y = 0\text{ m}$ in $5 \times 10^{-3}\text{ s}$ is (Assume parallel motion along the x-axis)

If the linear momentum is increased by 50%, then kinetic energy will increase by

A uniform metallic rod rotates about its perpendicular bisector with constant angular speed. If it is heated uniformly to raise its temperature slightly, then

Consider the situation shown in figure. A spring of spring constant 400 N/m is attached at one end to a wedge fixed rigidly with the horizontal part. A 40 g mass is released from rest while situated at a height 5 cm the curved track. The minimum deformation in the spring is nearly equal to (take g = 10 m/s²)

The force on a particle as the function of displacement x (in x-direction) is given by F = 10 + 0.5x. The work done corresponding to displacement of particle from x = 0 to x = 2 unit is

Assertion: Total energy is negative for a bound system. Reason: Potential energy of a bound system is negative and more than kinetic energy.

Assertion: In elastic collision, kinetic energy is conserved. Reason: Energy is always conserved.

Assertion: KE is conserved at every instant of (elastic) collision. Reason: No deformation of matter occurs in elastic collision.

A block of mass m is pulled along a horizontal surface by applying a force at an angle θ with the horizontal. If the block travels with a uniform velocity and has a displacement d and the coefficient of friction is μ, then the work done by the applied force is

A block of mass 10 kg is moving in x-direction with a constant speed of 10 m/s. It is subjected to a retarding force F = 0.1x joule/meter during its travel from x = 20 m to x = 30 m. Its final K.E. will be

A block of mass 10 kg is moving in x-direction with a constant speed of 10 m/s. It is subjected to a retarding force F = 0.1x joule/metre during its travel from x = 20 m to x = 30 m. Its final K.E. will be

Assertion: If momentum of a body increases by 50%, its kinetic energy will increase by 125%. Reason: Kinetic energy is proportional to square of velocity.

A force \( F \) acting on an object varies with distance \( x \) as shown in the figure. The force is in N and \( x \) in m. The work done by the force in moving the object from \( x = 0 \) to \( x = 6 \) m is

A body of mass 5 kg moving with a speed of 1.5 m/s on a horizontal smooth surface collides with a nearly weightless spring of force constant k = 5 N/m. The maximum compression of the spring would be

A body is moved along a straight line by a machine delivering constant power. The distance travelled by the body in time t is proportional to

A ball is bouncing down a flight of stairs. The coefficient of restitution is e. The height of each step is d and the ball descends one step each bounce. After each bounce it rebounds to a height h above the next lower step. The height is large enough compared with the width of step so that the impacts are effectively head-on. Find the relationship between h and d.

For inelastic collision between two spherical rigid bodies

A block of mass 10 kg is moving in x-direction with a constant speed of 10 m/sec. It is subjected to a retarding force F = 0.1x joule/meter during its travel from x = 20 meter to x = 30 meter. Its final kinetic energy will be

Energy required to break one bond in DNA is approximately

A bomb of mass 3.0 kg explodes in air into two pieces of masses 2.0 kg and 1.0 kg. The smaller mass goes at a speed of 80 m/s. The total energy imparted to the two fragments is

A neutron makes a head-on elastic collision with a stationary deuteron. The fractional energy loss of the neutron in the collision is

Assertion : In an elastic collision of two billiard balls, the total kinetic energy is conserved during the short time of oscillation of the balls (i.e. when they are in contact). Reason : Energy spent against friction does not follow the law of conservation of energy.

Maximum energy transfer for an elastic collision will occur if one body is at rest with

Two particles are seen to collide and move jointly together after the collision. During such a collision, for the total system,

Assertion (A) : Work done in uniform circular motion is zero. Reason (R) : Force is always directed along displacement.

A body is allowed to slide down a frictionless track freely under gravity. The track ends in a semicircular shaped part of diameter D. What should be the height (minimum) from which the body must fall so that it completes the circle.

A body of mass 5 kg has momentum of 10 kg m/s. When a force of 0.2 N is applied on it for 10 seconds, what is the change in its kinetic energy?

Which of the following is true?

Two spheres of equal mass collide with the collision being absolutely elastic but not central. Then the angle between the velocities (θ) must be