To determine the potential energy in a stretched spring, we have the formula for elastic potential energy: U = (1/2) k x² Where: - U is the potential energy - k is the spring constant - x is the extension (or compression) of the spring from its equilibrium position Given: - When the spring is stretcRead more
To determine the potential energy in a stretched spring, we have the formula for elastic potential energy:
U = (1/2) k x²
Where:
– U is the potential energy
– k is the spring constant
– x is the extension (or compression) of the spring from its equilibrium position
Given:
– When the spring is stretched by x₁ = 2 cm:
U = (1/2) k (2 cm)² = (1/2) k (0.02 m)²
– When the spring is stretched by x₂ = 8 cm:
U’ = (1/2) k (8 cm)² = (1/2) k (0.08 m)²
Step 1: Calculate the ratio of potential energies
We are required to find the ratio of the potential energies when stretched by 8 cm compared to when stretched by 2 cm:
(U’) / (U) = [(1/2) k (0.08 m)²] / [(1/2) k (0.02 m)²] = [(0.08)²] / [(0.02)²]
To find the resultant velocity after the collision, we can use the principle of conservation of momentum. Momentum before collision is equal to the momentum after collision. Let: - Mass of moving body = m - Velocity of moving body = v₁ = 3 km/h - Mass of resting body = 2m - Velocity of resting bodyRead more
To find the resultant velocity after the collision, we can use the principle of conservation of momentum. Momentum before collision is equal to the momentum after collision.
Let:
– Mass of moving body = m
– Velocity of moving body = v₁ = 3 km/h
– Mass of resting body = 2m
– Velocity of resting body = v₂ = 0 km/h
– Combined mass after the collision = m + 2m = 3m
– Combined velocity after the collision = vբ
Total momentum before collision is:
Momentum Initial = mv₁ + (2m)v₂ = m(3) + (2m)(0) = 3m
Total momentum after collision:
Final Momentum = (3m)vբ
Put the initial momentum equal to final momentum:
3m = 3mvբ
Now solve for բv:
բv = 3m / 3m = 1 km/h
Thus, the resultant velocity after collision is 1 km/hour.
In physics, work is defined as the transfer of energy through force applied over a distance. It can be classified into three types: positive work, negative work, and zero work. Positive Work: Positive work occurs when the force applied to an object and the displacement of that object are in the sameRead more
In physics, work is defined as the transfer of energy through force applied over a distance. It can be classified into three types: positive work, negative work, and zero work.
Positive Work:
Positive work occurs when the force applied to an object and the displacement of that object are in the same direction. This means that energy is being transferred to the object, causing it to move.
Example: When you push a box on the ground in the direction of the push, then the work done is positive since the force and the displacement are in the same direction.
Negative Work:
Negative work occurs when the force applied to an object and its displacement are in opposite directions. In this case, energy is being taken away from the object, slowing it down or causing it to lose energy.
Example: When brakes are applied to a moving car, the force exerted by brakes acts opposite to the direction of car motion. Work done by brakes will be negative since it decreases the kinetic energy of the car.
Zero Work:
This means zero work is when the displacement is in the direction of an applied force, or the displacement is perpendicular to the force. Thus, even if a force may be applied, no energy will have been transferred to the object in that direction.
For instance, when carrying a heavy bag at constant velocity while walking, the upward force applied by the hand to the bag does not contribute any work to the bag along the horizontal direction of travel. The displacement of the bag is perpendicular to the force due to gravity on the bag. If one exerts effort on pushing against a wall without causing the wall to budge, the work performed is zero.
The relationship between a satellite's kinetic energy and potential energy is one that is unique for a stabilized Earth orbit. The motion of the satellite is controlled by gravitation, which gives the centripetal force to keep it curved along the path around the earth. In this system, the satelliteRead more
The relationship between a satellite’s kinetic energy and potential energy is one that is unique for a stabilized Earth orbit. The motion of the satellite is controlled by gravitation, which gives the centripetal force to keep it curved along the path around the earth.
In this system, the satellite contains two types of energy: one is kinetic due to its motion and the other is potential because of the Earth’s gravitational pull. A prominent result of orbital mechanics is that the kinetic energy of the satellite is always equal in magnitude to half of its potential energy but has opposite signs. This means that the ratio of kinetic energy to potential energy is always 1/2.
This bond ensures the satellite remains in orbit. The total energy of the system is negative, since it is the sum of the kinetic and potential energy, representing the bound state of the satellite. This concept is applicable for circular or elliptical orbits and plays a critical role in explaining the dynamics of satellites. In fact, this explains why the satellites remain stable in their orbits and do not drift away or spiral inward. The balance of these forms of energy is preserved.
For points located outside the Earth, the acceleration due to gravity decreases with distance. It is inversely proportional to the square of the distance from the center of the Earth, meaning as one moves further away, gravity becomes weaker. Specifically, this means that the gravitational force felRead more
For points located outside the Earth, the acceleration due to gravity decreases with distance. It is inversely proportional to the square of the distance from the center of the Earth, meaning as one moves further away, gravity becomes weaker. Specifically, this means that the gravitational force felt at a height above Earth’s surface diminishes with increasing distance from the center.
In contrast, for points inside the Earth, gravity behaves differently. The acceleration due to gravity decreases linearly as one moves closer to the center. This indicates that the gravitational pull inside the Earth is directly related to the distance from the center.
For points lying onside the earth (r > R)
gₕ/g = R²/(R + h)² = R²/r² or gₕ = gR²/r²
gₕ ∝ 1/r²
For points lying inside the earth (r < R)
gₔ = g(R – d)/R = gʳ/R or gₔ = gʳ/R
gₔ ∝ r
The potential energy of a long spring when stretched by 2 cm is U. If the spring is stretched by 8 cm, the potential energy stored in it is
To determine the potential energy in a stretched spring, we have the formula for elastic potential energy: U = (1/2) k x² Where: - U is the potential energy - k is the spring constant - x is the extension (or compression) of the spring from its equilibrium position Given: - When the spring is stretcRead more
To determine the potential energy in a stretched spring, we have the formula for elastic potential energy:
U = (1/2) k x²
Where:
– U is the potential energy
– k is the spring constant
– x is the extension (or compression) of the spring from its equilibrium position
Given:
– When the spring is stretched by x₁ = 2 cm:
U = (1/2) k (2 cm)² = (1/2) k (0.02 m)²
– When the spring is stretched by x₂ = 8 cm:
U’ = (1/2) k (8 cm)² = (1/2) k (0.08 m)²
Step 1: Calculate the ratio of potential energies
We are required to find the ratio of the potential energies when stretched by 8 cm compared to when stretched by 2 cm:
(U’) / (U) = [(1/2) k (0.08 m)²] / [(1/2) k (0.02 m)²] = [(0.08)²] / [(0.02)²]
Step 2: Simplify the ratio
Compute the squares:
(U’) / (U) = [(0.08)²] / [(0.02)²] = [0.0064] / [0.0004] = 16
Conclusion
The potential energy in the spring stretched by 8 cm is given by:
U’ = 16 U
For more solutions:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-5/
A moving body of mass m and velocity 3 km/h collides with a rest body of mass 2 m and stricks to it. Now the combined mass starts to move. What will be the combined velocity?
To find the resultant velocity after the collision, we can use the principle of conservation of momentum. Momentum before collision is equal to the momentum after collision. Let: - Mass of moving body = m - Velocity of moving body = v₁ = 3 km/h - Mass of resting body = 2m - Velocity of resting bodyRead more
To find the resultant velocity after the collision, we can use the principle of conservation of momentum. Momentum before collision is equal to the momentum after collision.
Let:
– Mass of moving body = m
– Velocity of moving body = v₁ = 3 km/h
– Mass of resting body = 2m
– Velocity of resting body = v₂ = 0 km/h
– Combined mass after the collision = m + 2m = 3m
– Combined velocity after the collision = vբ
Total momentum before collision is:
Momentum Initial = mv₁ + (2m)v₂ = m(3) + (2m)(0) = 3m
Total momentum after collision:
Final Momentum = (3m)vբ
Put the initial momentum equal to final momentum:
3m = 3mvբ
Now solve for բv:
բv = 3m / 3m = 1 km/h
Thus, the resultant velocity after collision is 1 km/hour.
Click for more:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-5/
What is meant by positive work, negative work and zero works? Give examples of each type.
In physics, work is defined as the transfer of energy through force applied over a distance. It can be classified into three types: positive work, negative work, and zero work. Positive Work: Positive work occurs when the force applied to an object and the displacement of that object are in the sameRead more
In physics, work is defined as the transfer of energy through force applied over a distance. It can be classified into three types: positive work, negative work, and zero work.
Positive Work:
Positive work occurs when the force applied to an object and the displacement of that object are in the same direction. This means that energy is being transferred to the object, causing it to move.
Example: When you push a box on the ground in the direction of the push, then the work done is positive since the force and the displacement are in the same direction.
Negative Work:
Negative work occurs when the force applied to an object and its displacement are in opposite directions. In this case, energy is being taken away from the object, slowing it down or causing it to lose energy.
Example: When brakes are applied to a moving car, the force exerted by brakes acts opposite to the direction of car motion. Work done by brakes will be negative since it decreases the kinetic energy of the car.
Zero Work:
This means zero work is when the displacement is in the direction of an applied force, or the displacement is perpendicular to the force. Thus, even if a force may be applied, no energy will have been transferred to the object in that direction.
For instance, when carrying a heavy bag at constant velocity while walking, the upward force applied by the hand to the bag does not contribute any work to the bag along the horizontal direction of travel. The displacement of the bag is perpendicular to the force due to gravity on the bag. If one exerts effort on pushing against a wall without causing the wall to budge, the work performed is zero.
For more info:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-5/
For a satellite moving in an orbit around the earth, the ratio of kinetic energy to potential energy is
The relationship between a satellite's kinetic energy and potential energy is one that is unique for a stabilized Earth orbit. The motion of the satellite is controlled by gravitation, which gives the centripetal force to keep it curved along the path around the earth. In this system, the satelliteRead more
The relationship between a satellite’s kinetic energy and potential energy is one that is unique for a stabilized Earth orbit. The motion of the satellite is controlled by gravitation, which gives the centripetal force to keep it curved along the path around the earth.
In this system, the satellite contains two types of energy: one is kinetic due to its motion and the other is potential because of the Earth’s gravitational pull. A prominent result of orbital mechanics is that the kinetic energy of the satellite is always equal in magnitude to half of its potential energy but has opposite signs. This means that the ratio of kinetic energy to potential energy is always 1/2.
This bond ensures the satellite remains in orbit. The total energy of the system is negative, since it is the sum of the kinetic and potential energy, representing the bound state of the satellite. This concept is applicable for circular or elliptical orbits and plays a critical role in explaining the dynamics of satellites. In fact, this explains why the satellites remain stable in their orbits and do not drift away or spiral inward. The balance of these forms of energy is preserved.
Click here for more: – https://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-6/
See lessDraw a graph showing the variation of acceleration dur to gravity g with distance r from the centre of the earth.
For points located outside the Earth, the acceleration due to gravity decreases with distance. It is inversely proportional to the square of the distance from the center of the Earth, meaning as one moves further away, gravity becomes weaker. Specifically, this means that the gravitational force felRead more
For points located outside the Earth, the acceleration due to gravity decreases with distance. It is inversely proportional to the square of the distance from the center of the Earth, meaning as one moves further away, gravity becomes weaker. Specifically, this means that the gravitational force felt at a height above Earth’s surface diminishes with increasing distance from the center.
In contrast, for points inside the Earth, gravity behaves differently. The acceleration due to gravity decreases linearly as one moves closer to the center. This indicates that the gravitational pull inside the Earth is directly related to the distance from the center.
For points lying onside the earth (r > R)
gₕ/g = R²/(R + h)² = R²/r² or gₕ = gR²/r²
gₕ ∝ 1/r²
For points lying inside the earth (r < R)
See lessgₔ = g(R – d)/R = gʳ/R or gₔ = gʳ/R
gₔ ∝ r