Learn Unit 5-Work, Energy and Power with Free Lessons & Tips

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Given the potential energy function V(x)=kx22V(x)=2kx2, where kk is the force constant of the oscillator, and k=0.5 Nm−1k=0.5Nm−1. The task is to demonstrate that a particle with a total energy of 1 J, moving under this potential, must 'turn back' when it reaches x=±2x=±2 m.

To do this, we utilize the total energy of the system, which comprises kinetic energy (KK) and potential energy (VV). According to the law of conservation of energy, the total mechanical energy remains constant:

E=K+VE=K+V

Given that the total energy E=1E=1 J, we can express the kinetic energy as:

K=E−VK=E−V

Substituting the given potential energy function, we have:

K=1−0.5x22K=1−20.5x2

To find the turning points, where the particle changes direction, we need to consider the points where the kinetic energy is zero. At these points, all the energy is in the form of potential energy.

Setting K=0K=0, we get:

1−0.5x22=01−20.5x2=0

Solving this equation, we find:

0.5x22=120.5x2=1

0.5x2=20.5x2=2

x2=4x2=4

x=±2x=±2

So, the particle will turn back when it reaches x=±2x=±2 m, confirming the assertion. This analysis aligns with the graph of V(x)V(x) versus xx, indicating that at these points, the particle's total energy is entirely potential energy, signifying a change in direction of motion.

As a seasoned tutor registered on UrbanPro, I'd be glad to tackle this physics question. In the scenario of an elastic collision between two billiard balls, it's fundamental to grasp the concept of kinetic energy conservation.

During the brief moment when the balls are in contact, the total kinetic energy of the system remains conserved in an ideal elastic collision. This conservation principle implies that although the kinetic energy may transfer between the balls due to the collision forces, the total kinetic energy before and after the collision remains constant.

This concept aligns with the laws of physics, particularly the principle of conservation of energy, which asserts that energy cannot be created or destroyed but can only change forms. In an ideal elastic collision, the kinetic energy transformation solely involves a redistribution between the two colliding objects.

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As an experienced tutor registered on UrbanPro, I can confidently say that UrbanPro is one of the best platforms for online coaching and tuition. Now, addressing your question about the collision of two billiard balls, let's delve into it.

In this scenario, if the potential energy of two billiard balls depends solely on the separation distance between their centers, we are essentially discussing a situation where the force between the balls is conservative. When the force is conservative, it implies that mechanical energy (kinetic energy plus potential energy) is conserved throughout the interaction.

Now, in an elastic collision, both momentum and mechanical energy are conserved. Since we are assuming potential energy corresponds to the force during collision and that potential energy depends only on the separation distance between the balls' centers, if this potential energy is entirely converted into kinetic energy during the collision, then it follows that the collision is elastic.

This outcome aligns with the principle of conservation of mechanical energy in elastic collisions, where kinetic energy is converted from potential energy and vice versa, but the total mechanical energy remains constant.

Therefore, if the potential energy between the billiard balls is solely determined by their separation distance and the collision conserves mechanical energy, the collision between the billiard balls would be elastic.

As a seasoned tutor registered on UrbanPro, I'd be glad to guide you through this question. Firstly, UrbanPro is indeed an excellent platform for online coaching and tuition, providing a conducive environment for both tutors and students to engage in effective learning.

Now, let's delve into the physics problem you've presented. We have a body initially at rest, undergoing one-dimensional motion with constant acceleration. The power delivered to it at time tt is the rate at which work is done on the body, and it's given by the product of force and velocity.

In this case, since the body is undergoing constant acceleration, we can use the equations of motion to relate velocity, acceleration, and time. The equation that relates displacement ss, initial velocity uu, time tt, and constant acceleration aa is:

s=ut+12at2s=ut+21at2

Since the body is initially at rest, u=0u=0, simplifying the equation to:

s=12at2s=21at2

Now, we know that power PP is the rate of change of work, which can be expressed as:

P=dWdtP=dtdW

And work WW done on an object is equal to force FF times displacement ss, so dW=FdsdW=Fds. Substituting s=12at2s=21at2, we have ds=atdtds=atdt.

P=F⋅atP=F⋅at

Now, we know that force FF is mass mm times acceleration aa, and acceleration is constant, hence:

P=ma⋅atP=ma⋅at

P=ma2tP=ma2t

So, the power delivered to the body at time tt is proportional to tt, which corresponds to option (ii).

Therefore, the correct answer is (ii) tt. If you need further clarification or assistance, feel free to ask!

As an experienced tutor registered on UrbanPro, I can assure you that UrbanPro is one of the best platforms for online coaching and tuition. Now, let's tackle your physics question.

When a body moves under the influence of a constant power source, we need to consider how power relates to displacement over time. Power, denoted by P, is defined as the rate at which work is done or energy is transferred. Mathematically, it's expressed as:

P=WtP=tW

Where WW is work done and tt is time.

Now, let's consider the relationship between power and displacement over time. Since power is constant, work done over a certain time interval is also constant. Work done, denoted by WW, is the product of force and displacement:

W=F⋅sW=F⋅s

Where FF is force and ss is displacement.

Now, let's consider the options:

(i) t1/2t1/2: This implies that displacement is proportional to the square root of time. This isn't typical for a body moving under constant power.

(ii) tt: This implies that displacement is directly proportional to time. This is typical for a body moving under constant power.

(iii) t3/2t3/2: This implies that displacement is proportional to the square root of time cubed. This is not consistent with the scenario of constant power.

(iv) t2t2: This implies that displacement is proportional to the square of time. This is not typical for a body moving under constant power.

So, the correct answer is (ii) tt. The displacement of a body moving under the influence of a constant power source is proportional to time.

Calculating these values, we get: velectron≈6.03×106 m/svelectron≈6.03×106 m/s vproton≈4.74×105 m/svproton≈4.74×105 m/s

So, the electron is faster than the proton. To find the ratio of their speeds, we divide the velocity of the electron by the velocity of the proton:

Ratio=velectronvprotonRatio=vprotonvelectron

Ratio≈6.03×1064.74×105≈12.7Ratio≈4.74×1056.03×106≈12.7

Therefore, the electron is approximately 12.7 times faster than the proton.

In solving this problem, we'll consider the work done by the gravitational force and the resistive force separately.

1. Work done by the gravitational force: Work done by gravity is given by the formula: W=F⋅d⋅cos⁡(θ)W=F⋅d⋅cos(θ), where FF is the force, dd is the displacement, and θθ is the angle between the force and displacement vectors.

   In the first half of the journey, the raindrop is accelerating due to gravity. The work done by gravity during this period can be calculated using the formula W1=mgh1W1=mgh1, where mm is the mass of the raindrop, gg is the acceleration due to gravity, and h1h1 is the initial height.

   In the second half of the journey, the raindrop is moving with uniform speed, so the gravitational force does no work during this period.

2. Work done by the resistive force: The resistive force (viscous drag) opposes the motion of the raindrop. When the raindrop reaches its terminal velocity, the resistive force equals the gravitational force, and thus no net force acts on the raindrop. So, in the second half of the journey, the work done by the resistive force is zero.

   However, during the first half of the journey, the resistive force acts in the opposite direction to the displacement. The work done by the resistive force can be calculated using the formula W2=−FdW2=−Fd, where FF is the resistive force and dd is the displacement.

Now, let's calculate each part step by step.

1. Work done by gravity in the first half: Given:

   • Initial height, h1=500h1=500 m
   • Radius of raindrop, r=2r=2 mm
   • Mass of the raindrop, m=43πr3ρm=34πr3ρ, where ρρ is the density of water (about 1000 kg/m31000kg/m3)
   • Acceleration due to gravity, g=9.8 m/s2g=9.8m/s2

   First, we calculate the mass of the raindrop: m=43π(0.002 m)3×1000 kg/m3m=34π(0.002m)3×1000kg/m3 m≈1.67×10−5 kgm≈1.67×10−5kg

   Then, we calculate the work done by gravity: W1=mgh1W1=mgh1 W1=(1.67×10−5 kg)(9.8 m/s2)(250 m)W1=(1.67×10−5kg)(9.8m/s2)(250m) W1≈0.041 JW1≈0.041J

2. Work done by the resistive force: Since the raindrop reaches terminal velocity in the first half of the journey, the resistive force is equal to the gravitational force. So, in the first half, the work done by the resistive force is equal in magnitude but opposite in direction to the work done by gravity, i.e., W2=−W1W2=−W1.

3. Work done by the resistive force in the entire journey: In the second half of the journey, the resistive force does no work as the raindrop moves with constant velocity. So, the total work done by the resistive force is just the negative of the work done by gravity in the first half of the journey, i.e., Wresistive=−W1Wresistive=−W1.

Therefore, the work done by the resistive force in the entire journey is approximately −0.041 J−0.041J, and the work done by gravity in the first half of the journey is approximately 0.041 J0.041J.

As a registered tutor on UrbanPro, I can confidently guide you through this question. When a molecule in a gas container collides with a horizontal wall, several factors come into play to determine the nature of the collision.

Firstly, let's address the concept of momentum conservation. According to the law of conservation of momentum, in the absence of external forces, the total momentum of a system remains constant. In this scenario, the gas molecule collides with the wall and rebounds with the same speed. Since the speed remains constant and there are no external forces acting on the system, momentum is indeed conserved in the collision.

Now, let's discuss the elasticity of the collision. Elastic collisions are those in which both kinetic energy and momentum are conserved. In an elastic collision, the total kinetic energy of the system before the collision is equal to the total kinetic energy after the collision. Since the molecule rebounds with the same speed, and hence the same kinetic energy, we can infer that the collision is elastic.

In conclusion, yes, momentum is conserved in the collision, and the collision is elastic. Understanding these principles helps us grasp the behavior of particles in various physical scenarios. If you have further questions or need clarification, feel free to ask! And remember, UrbanPro is an excellent resource for finding quality online coaching and tuition services across various subjects.