#### A car of mass 250 kg changes its velocity from 10 m/s to 20 m/s. Then, the change in its kinetic energy is ____ J.

As, mass, m = 250 kg; initial velocity, v_{1} = 10 m/s; final velocity, v_{2} = 20 m/s
Change in kinetic energy, ΔKE = ½ m (v^{2}_{2} – v^{2}_{1})
ΔKE= ½ x 250 x (400-100)
ΔKE=37500 J
Change in kinetic energy is 37500 J.

#### To describe work done on an object, we require:

To describe work done on an object, we require: The magnitude of the net force acting on the object, its initial position and its final position.

#### Sound energy is converted into electrical energy by ____.

Sound energy is converted into electrical energy by a microphone.

#### Work done in raising a box on to a platform depends on ____.

Work done in raising a box on to a platform depends on the height to which it is raised. When a body is raised to a height then work done is equal to the potential energy.

#### The unit Nm is equivalent to ____.

Unit Nm is equivalent to J.

#### Work done by the gravitational pull of the Sun on the Earth is maximum in the month of July.

Gravitational pull of the Sun on the Earth plays the role of centripetal force. So, at any instant gravitational force and displacement of Earth are perpendicular to each other. Work done, W = FScosθ here θ = 90° and cos90° = 0, so, W = 0 The work done by the gravitational pull of the Sun on the Earth is zero.

#### A truck and a car are moving on a smooth, level road such that the kinetic energy associated with them is same. When the same brake force is applied on both of them simultaneously, the truck will cover a greater distance before it stops.

Work done is equal to change in kinetic energy. So, work done by the brake force is same. Work done, W = Fs; as work done and force is same, both truck and car travel same distance.

#### A body in motion comes to rest when force is applied on it. The work done by the force on the body is ____.

Force and displacements are in opposite direction. A body in motion comes to rest when force is applied on it. Hence, the work done by the force on the body is negative.

#### Work done by the gravitational force on a body is ____.

When a body is carried along the direction of the gravitational force, the work done is positive and vice-versa. When the displacement is made perpendicular to the direction of the gravitational force, the work done is zero.

#### A body is moved from A to B in three different paths in conservative force field. The work done will be maximum when the body is moved along the longest path.

Work done by the conservative force does not depend upon the path followed by the body. It depends upon the displacement covered from initial position to final position.

#### When the pendulum bob is in the mean position, total energy will be minimum.

When the pendulum bob is in the mean position, the total energy will remain unchanged. Total energy is the sum of KE and PE. At mean position, the KE is maximum and PE is zero. Hence, the total energy remains unchanged.

#### Work done by the electrostatic force on an electron revolving around the nucleus of hydrogen is ____J.

The electrostatic force and displacement of an electron are mutually perpendicular to each other. Hence, the work done by the electrostatic force on an electron revolving around the nucleus of hydrogen is zero.

#### Energy is a scalar.

Energy is a scalar as it has only magnitude.

#### Potential and kinetic energies are scalar quantities.

Potential and kinetic energies are scalar quantities.

#### Work done on a body is a vector quantity.

Work done on a body is a scalar quantity.

#### When force is applied on a body, work done by it is always positive.

When force is applied on a body, work done by it can be positive, negative or zero.

#### A watch spring in the wounded position has both kinetic as well as potential energies.

Potential energy is stored in the spring by change in configuration of the body.

#### Work is measured as a product of ____.

Work done is given by the product of force and displacement, represented by, Work = force × displacement.

#### When the force applied and the displacement of the body is inclined at 90°, with each other, the work done is ____.

Work done = FScosθ As, cos90° = 0 Therefore, work done is zero.

#### A bird flying in the sky has only kinetic energy.

Flying bird has both velocity as well as altitude so it has both KE and PE.

#### If the momentum of a body is increased to 5 times the initial value, then the KE is____ times its initial value.

According to the problem, initial KE = p^{2}/2m
Final KE = (5p)^{2}/2m
Final KE = 25p^{2}/2m
KE increases 25 times.

#### A labour carrying a suitcase on his head moves from rest on a horizontal road to another point and finally comes to rest. Then, the work done by gravity is ____.

As, W = FScosθ As the angle between the force applied by gravity and displacement, θ = 90° Hence, work done is zero.

#### In the SI system, the unit of potential energy, PE, is ____.

In the SI system, the unit of potential energy, PE, is joule.

#### 1 kJ = ____ J.

1 kJ = 10^{3} J

#### kg m^{2}/s^{2} represents the unit of ____.

Work done, W = FScosθ
SI unit of work is N m (or) kg m^{2}/s^{2}
and torque = r × F
SI unit of torque is N m (or) kg m^{2}/s^{2}
Hence, work done and torque have same unit.

#### When a body is released from a height the kinetic energy of the body is gradually changed into potential energy.

When a body is released from a height the potential energy of the body is gradually changed into kinetic energy. In this case total mechanical energy is conserved or constant. Total energy = KE + PE

#### A body is dropped from a certain height. When it is halfway down, it possesses ____.

When the body is halfway down, it possesses both kinetic and potential energy.

#### The velocity of a body of mass 100 g having kinetic energy of 20 J is ____ m/s.

Given, m = 100 g = 0.1 kg, KE = 20 J
but, KE = 1/2mv^{2}
⇒ 20 = 1/2 × 0.1 × v^{2}
∴ v = 20 m/s
Therefore, the velocity of the body of mass 100 g having kinetic energy of 20 J is 20 m/s.

#### Work done by force acting on an object is equal to the magnitude of the force multiplied by the displacement of the object in the direction of the force.

Work done by force acting on an object is equal to the magnitude of the force multiplied by the displacement of the object in the direction of the force.

Share your Results: