Class 12 Exam  >  Class 12 Questions  >  A block Of mass 1 kg falls freely on a spring... Start Learning for Free
A block Of mass 1 kg falls freely on a spring from a height Of 20 CM as shown in find the compression in the spring if it's force constant is 10>3n/m?
Most Upvoted Answer
A block Of mass 1 kg falls freely on a spring from a height Of 20 CM a...
Community Answer
A block Of mass 1 kg falls freely on a spring from a height Of 20 CM a...
**Solution:**

When a block of mass 1 kg falls freely from a height of 20 cm on a spring, the potential energy of the block is converted into the spring potential energy.

We can use the principle of conservation of energy to find the maximum compression in the spring. According to this principle, the total energy of a closed system remains constant.

The total energy of the system is given by:

Total energy = Potential energy + Kinetic energy + Spring potential energy

Initially, the block has only potential energy, given by:

Potential energy = mgh

where m is the mass of the block, g is the acceleration due to gravity, and h is the height from which the block is dropped.

Potential energy = 1 x 9.8 x 0.2 = 1.96 J

As the block falls, its potential energy is converted into kinetic energy, given by:

Kinetic energy = (1/2)mv^2

where m is the mass of the block and v is its velocity just before it hits the spring.

The velocity of the block just before it hits the spring can be found from the conservation of energy equation:

Total energy = Potential energy + Kinetic energy + Spring potential energy

At the maximum compression of the spring, the kinetic energy of the block is zero. Therefore, the total energy of the system is equal to the sum of its potential energy and spring potential energy.

Total energy = Potential energy + Spring potential energy

The spring potential energy is given by:

Spring potential energy = (1/2)kx^2

where k is the force constant of the spring and x is the compression in the spring.

Substituting the values of potential energy, spring potential energy, and force constant, we get:

1.96 J = (1/2)kx^2

x^2 = (2 x 1.96) / k

x^2 = 0.00392

x = 0.0626 m

Therefore, the maximum compression in the spring is 6.26 cm.

**Explanation:**

The solution is based on the principle of conservation of energy. The potential energy of the block is converted into kinetic energy just before it hits the spring. At the maximum compression of the spring, the kinetic energy is zero, and all the energy is stored in the spring potential energy. This spring potential energy is given by (1/2)kx^2, where k is the force constant of the spring and x is the compression in the spring. By equating the total energy of the system to the sum of its potential and spring potential energy, we can find the maximum compression in the spring.
Explore Courses for Class 12 exam

Similar Class 12 Doubts

Read the following text and answer the following questions on the basis of the same: Roget’s spiral: Magnetic effects are generally smaller than electric effects. As a consequence, the force between currents is rather small, because of the smallness of the factor μ. Hence, it is difficult to demonstrate attraction or repulsion between currents. Thus, for 5 A current in each wire at a separation of 1 cm, the force per metre would be 5 × 10–4 N, which is about 50 mg weight. It would be like pulling a wire by a string going over a pulley to which a 50 mg weight is attached. The displacement of the wire would be quite unnoticeable. With the use of a soft spring, we can increase the effective length of the parallel current and by using mercury, we can make the displacement of even a few mm observable very dramatically. You will also need a constant-current supply giving a constant current of about 5 A. Take a soft spring whose natural period of oscillations is about 0.5–1 s. Hang it vertically and attach a pointed tip to its lower end, as shown in the figure here. Take some mercury in a dish and adjust the spring such that the tip is just above the mercury surface. Take the DC current source, connect one of its terminals to the upper end of the spring and dip the other terminal in mercury. If the tip of the spring touches mercury, the circuit is completed through mercury. Let the DC source be put off to begin with. Let the tip be adjusted so that it just touches the mercury surface. Switch on the constant current supply and watch the fascinating outcome. The spring shrinks with a jerk, the tip comes out of mercury (just by a mm or so), the circuit is broken, the current stops, the spring relaxes and tries to come back to its original position, the tip again touches mercury establishing a current in the circuit and the cycle continues with tick, tick, tick,...The force 10–3 N,is equivalent to

Read the following text and answer the following questions on the basis of the same: Roget’s spiral: Magnetic effects are generally smaller than electric effects. As a consequence, the force between currents is rather small, because of the smallness of the factor μ. Hence, it is difficult to demonstrate attraction or repulsion between currents. Thus, for 5 A current in each wire at a separation of 1 cm, the force per metre would be 5 × 10–4 N, which is about 50 mg weight. It would be like pulling a wire by a string going over a pulley to which a 50 mg weight is attached. The displacement of the wire would be quite unnoticeable. With the use of a soft spring, we can increase the effective length of the parallel current and by using mercury, we can make the displacement of even a few mm observable very dramatically. You will also need a constant-current supply giving a constant current of about 5 A. Take a soft spring whose natural period of oscillations is about 0.5–1 s. Hang it vertically and attach a pointed tip to its lower end, as shown in the figure here. Take some mercury in a dish and adjust the spring such that the tip is just above the mercury surface. Take the DC current source, connect one of its terminals to the upper end of the spring and dip the other terminal in mercury. If the tip of the spring touches mercury, the circuit is completed through mercury. Let the DC source be put off to begin with. Let the tip be adjusted so that it just touches the mercury surface. Switch on the constant current supply and watch the fascinating outcome. The spring shrinks with a jerk, the tip comes out of mercury (just by a mm or so), the circuit is broken, the current stops, the spring relaxes and tries to come back to its original position, the tip again touches mercury establishing a current in the circuit and the cycle continues with tick, tick, tick,...Why the spring shrinks in Roget’s spiral ?

Read the following text and answer the following questions on the basis of the same:Roget’s spiral: Magnetic effects are generally smaller than electric effects. As a consequence, the force between currents is rather small, because of the smallness of the factor μ. Hence, it is difficult to demonstrate attraction or repulsion between currents. Thus, for 5 A current in each wire at a separation of 1 cm, the force per metre would be 5 × 10–4 N, which is about 50 mg weight. It would be like pulling a wire by a string going over a pulley to which a 50 mg weight is attached. The displacement of the wire would be quite unnoticeable. With the use of a soft spring, we can increase the effective length of the parallel current and by using mercury, we can make the displacement of even a few mm observable very dramatically. You will also need a constant-current supply giving a constant current of about 5 A. Take a soft spring whose natural period of oscillations is about 0.5–1 s. Hang it vertically and attach a pointed tip to its lower end, as shown in the figure here. Take some mercury in a dish and adjust the spring such that the tip is just above the mercury surface. Take the DC current source, connect one of its terminals to the upper end of the spring and dip the other terminal in mercury. If the tip of the spring touches mercury, the circuit is completed through mercury. Let the DC source be put off to begin with. Let the tip be adjusted so that it just touches the mercury surface. Switch on the constant current supply and watch the fascinating outcome. The spring shrinks with a jerk, the tip comes out of mercury (just by a mm or so), the circuit is broken, the current stops, the spring relaxes and tries to come back to its original position, the tip again touches mercury establishing a current in the circuit and the cycle continues with tick, tick, tick,...Magnetic effects

Read the following text and answer the following questions on the basis of the same: Roget’s spiral: Magnetic effects are generally smaller than electric effects. As a consequence, the force between currents is rather small, because of the smallness of the factor μ. Hence, it is difficult to demonstrate attraction or repulsion between currents. Thus, for 5 A current in each wire at a separation of 1 cm, the force per metre would be 5 × 10–4 N, which is about 50 mg weight. It would be like pulling a wire by a string going over a pulley to which a 50 mg weight is attached. The displacement of the wire would be quite unnoticeable. With the use of a soft spring, we can increase the effective length of the parallel current and by using mercury, we can make the displacement of even a few mm observable very dramatically. You will also need a constant-current supply giving a constant current of about 5 A. Take a soft spring whose natural period of oscillations is about 0.5–1 s. Hang it vertically and attach a pointed tip to its lower end, as shown in the figure here. Take some mercury in a dish and adjust the spring such that the tip is just above the mercury surface. Take the DC current source, connect one of its terminals to the upper end of the spring and dip the other terminal in mercury. If the tip of the spring touches mercury, the circuit is completed through mercury. Let the DC source be put off to begin with. Let the tip be adjusted so that it just touches the mercury surface. Switch on the constant current supply and watch the fascinating outcome. The spring shrinks with a jerk, the tip comes out of mercury (just by a mm or so), the circuit is broken, the current stops, the spring relaxes and tries to come back to its original position, the tip again touches mercury establishing a current in the circuit and the cycle continues with tick, tick, tick,...What are the main 3 components in a Roget’s spiral?

Read the following text and answer the following questions on the basis of the same: Roget’s spiral: Magnetic effects are generally smaller than electric effects. As a consequence, the force between currents is rather small, because of the smallness of the factor μ. Hence, it is difficult to demonstrate attraction or repulsion between currents. Thus, for 5 A current in each wire at a separation of 1 cm, the force per metre would be 5 × 10–4 N, which is about 50 mg weight. It would be like pulling a wire by a string going over a pulley to which a 50 mg weight is attached. The displacement of the wire would be quite unnoticeable. With the use of a soft spring, we can increase the effective length of the parallel current and by using mercury, we can make the displacement of even a few mm observable very dramatically. You will also need a constant-current supply giving a constant current of about 5 A. Take a soft spring whose natural period of oscillations is about 0.5–1 s. Hang it vertically and attach a pointed tip to its lower end, as shown in the figure here. Take some mercury in a dish and adjust the spring such that the tip is just above the mercury surface. Take the DC current source, connect one of its terminals to the upper end of the spring and dip the other terminal in mercury. If the tip of the spring touches mercury, the circuit is completed through mercury. Let the DC source be put off to begin with. Let the tip be adjusted so that it just touches the mercury surface. Switch on the constant current supply and watch the fascinating outcome. The spring shrinks with a jerk, the tip comes out of mercury (just by a mm or so), the circuit is broken, the current stops, the spring relaxes and tries to come back to its original position, the tip again touches mercury establishing a current in the circuit and the cycle continues with tick, tick, tick,...What else can be used instead of mercury in Roget’s spiral ?

A block Of mass 1 kg falls freely on a spring from a height Of 20 CM as shown in find the compression in the spring if it's force constant is 10>3n/m?
Question Description
A block Of mass 1 kg falls freely on a spring from a height Of 20 CM as shown in find the compression in the spring if it's force constant is 10>3n/m? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about A block Of mass 1 kg falls freely on a spring from a height Of 20 CM as shown in find the compression in the spring if it's force constant is 10>3n/m? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A block Of mass 1 kg falls freely on a spring from a height Of 20 CM as shown in find the compression in the spring if it's force constant is 10>3n/m?.
Solutions for A block Of mass 1 kg falls freely on a spring from a height Of 20 CM as shown in find the compression in the spring if it's force constant is 10>3n/m? in English & in Hindi are available as part of our courses for Class 12. Download more important topics, notes, lectures and mock test series for Class 12 Exam by signing up for free.
Here you can find the meaning of A block Of mass 1 kg falls freely on a spring from a height Of 20 CM as shown in find the compression in the spring if it's force constant is 10>3n/m? defined & explained in the simplest way possible. Besides giving the explanation of A block Of mass 1 kg falls freely on a spring from a height Of 20 CM as shown in find the compression in the spring if it's force constant is 10>3n/m?, a detailed solution for A block Of mass 1 kg falls freely on a spring from a height Of 20 CM as shown in find the compression in the spring if it's force constant is 10>3n/m? has been provided alongside types of A block Of mass 1 kg falls freely on a spring from a height Of 20 CM as shown in find the compression in the spring if it's force constant is 10>3n/m? theory, EduRev gives you an ample number of questions to practice A block Of mass 1 kg falls freely on a spring from a height Of 20 CM as shown in find the compression in the spring if it's force constant is 10>3n/m? tests, examples and also practice Class 12 tests.
Explore Courses for Class 12 exam
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev