What is the energy conservation principle in a two masses and pulley problem?

In summary, when the heavier mass hits the ground then its KE i.e kinetic energy is dissipated as sound and heat energy and it can be assumed to come to rest. Then the lighter mass experiences no tension since the string now will not remain taut as lighter mass will continue moving up under gravity acceleration to a max height where it's velocity will be zero.The string can only work in tension, not in compression.For that reason, the 8-kg mass "can't feel" that impact force that stopped the 14-kg mass.It only "feels" the sudden absence of the 14g force that was pulling it up, but it "feels" no resistance to continue moving up as a projectile as far as
  • #1
Vigorous
33
3
Homework Statement
See the masses in Figure (1) which start out at rest. (i) Find the velocity of the 14 kg mass just before it hits the ground. (ii) Find the maximum height reached by the 8 kg (and don’t worry about hitting the pulley). (iii) Find the fraction of mechanical energy left when the system finally comes to rest.
Relevant Equations
1) Newtons laws
2) Energy Conservation
i) I first analyzed the forces as soon as the 14 Kg is released. The aim of this step is to calculate the work done by the net force acting on the 14 kg mass to determine the change in kinetic energy.
T-14g=-14a
T-8g=8a
T=99.7 N a=6g/22 m/s^2
Since the net force is constant and does not vary with distance
K2-K1=F*distance travelled=-37.418*5=187.1 =1/2mv2^2
K1=0 Since the system starts at rest. solving for v2
v2=5.169 m/s

When the system was at rest, was there an external force pushing down on the 8 kg mass to counter what the tension force wants to do. Similarly, another external force acting on the 14 kg mass to hold it at 5m off the ground. At the point of release, these external forces are removed. I am not sure of my reasoning.
 

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  • #2
Consider that gravity is acting on the whole system at once, as both masses are linked by a string of constant length.
 
  • #3
Yes, absolutely. There had to be some upward external force holding the 14 kg mass or pushing the 8 kg mass down else heavier mass would move downwards. But this external force is of no consequence when the heavier mass starts moving down from rest since at that point of time the external force is not there.
 
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  • #4
The question ought to specify that the string is inelastic. This is important for parts ii and iii.
 
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  • #5
Lnewqban said:
Consider that gravity is acting on the whole system at once, as both masses are linked by a string of constant length.
i) The two external forces acting on the system are the weights of the two masses. Since 14g>8g the system will accelerate in the direction of the heavier mass. a=14g-8g/14+8=6g/22. T=14g-14(6g/22).
Energy of the system
Change in kinetic energy of the system is equal to the work done by the net force acting on the system. The initial kinetic energy is zero since the system is at rest, the final kinetic energy is 1/2(14+8)v^2. The net force is 6g. The displacement is 5 m.
1/2(14+8)v^2=6g*5 v=5.169 m/s
ii) The 14 kg lands so do we add the normal force by the ground in computing the external forces acting on the system.
 
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  • #6
haruspex said:
The question ought to specify that the string is inelastic. This is important for parts ii and iii.
From that assumption if the 14 kg mass moves down 5 m the 8kg will move up by the same 5 m. If the displacement is the same so is the velocity and the acceleration.
ii) The 14 Kg hits the floor, a normal ground force acts on to stop it from accelerating. That normal force is the difference between its weight and the tension of the rope. For the 8kg mass the forces acting on it is the pull of gravity (8g) and the tension of the rope pulling it upwards, why does it come to a halt as it reaches its maximum height there must be an additional force that decelerates it.
 
  • #7
Vigorous said:
From that assumption if the 14 kg mass moves down 5 m the 8kg will move up by the same 5 m. If the displacement is the same so is the velocity and the acceleration.
ii) The 14 Kg hits the floor, a normal ground force acts on to stop it from accelerating. That normal force is the difference between its weight and the tension of the rope. For the 8kg mass the forces acting on it is the pull of gravity (8g) and the tension of the rope pulling it upwards, why does it come to a halt as it reaches its maximum height there must be an additional force that decelerates it.
When heavier mass hits the ground then its KE i.e kinetic energy is dissipated as sound and heat energy and it can be assumed to come to rest. Then the lighter mass experiences no tension since the string now will not remain taut as lighter mass will continue moving up under gravity acceleration to a max height where it's velocity will be zero.
 
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  • #8
The string can only work in tension, not in compression.
For that reason, the 8-kg mass "can't feel" that impact force that stopped the 14-Kg mass.
It only "feels" the sudden absence of the 14g force that was pulling it up, but it "feels" no resistance to continue moving up as a projectile as far as gravity will allow it to go (its accumulated kinetic energy will be transformed into potential energy.
Then, the 8-kg mass will be in free-fall until impacting the ground, just like the 14-kg mass did a little before.

As the 14-kg mass does not affect the system after impact, we should not be concerned with the type of impact (elastic or not) or the forces involved in it.
Same for the 8-kg mass during and after impact.
 
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  • #9
Lnewqban said:
the sudden absence of the 14g force that was pulling it up
... or somewhat less than 14g kg.
Lnewqban said:
As the 14-kg mass does not affect the system after impact, we should not be concerned with the type of impact (elastic or not)
Ok, but part iii requires us to assume both floor and string are inelastic.
 
  • #10
When the 14 kg mass hits the floor why does the tension equal to zero? I understand why the tension after the impact cannot equal its value just before the impact then the 8 Kg mass would continue to accelerate upward. It would be speeding up and gaining height and would result in having more energy than what it started with. Why isn't the tension equal to some value less than the weight of the 8kg? Why does the normal due to the ground have to support all of the weight of the 14 kg and not share part of it with the tension force?

-8gh+8g(5)=K2-K1=-1/2(8)(5.169)^2
K2=0 at the maximum height
K1-kinetic energy of the 8 kg mass at 5m off the ground
 
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  • #11
Vigorous said:
When the 14 kg mass hits the floor why does the tension equal to zero?
The smaller mass has an upward velocity, so it will continue up for a distance. The larger mass has stopped descending. Therefore the string is slack and does not exert any force.
Or are you thinking that the tension could be zero on one side and not on the other?
 
  • #12
haruspex said:
... or somewhat less than 14g kg.
...
I stand corrected.
The force pulling the 8-kg mass upwards is not 14g.
Thank you, haruspex.
 
  • #13
Vigorous said:
When the 14 kg mass hits the floor why does the tension equal to zero?
When a truck is towing a disabled car using a rope or a chain, what happens if the truck driver suddenly applies the brakes?
 
  • #14
haruspex said:
are you thinking that the tension could be zero on one side and not on the other?
No the tension on both sides should be equal because there cannot be a net force acting on a massless object.

@Lnewqban The rope cannot resist compression and will bend so if the truck is moving on a frictionless road the disabled car and the rope will detach from the truck. However, if the car was moving on a road with friction then friction would provide the deceleration of the disabled car.

The last part of this question the system comes to rest at two instants when the 8kg reaches maximum height and when it drops under free fall and hits the ground.
In calculating the initial energy of the 14 kg why is it wrong when I compute the change in kinetic energy between two times (when the 14 kg was at 5 m off the ground and when it hits the ground) to state the following
K(h=0)-K(h=5)=0(T-14g)-5(T-14g)
K(h=0)-0(T-14g)=5(T-14g)+K(h=5)=E(initial energy of the system)
 
  • #15
Vigorous said:
there cannot be a net force acting on a massless object.
Or, in this case, a net torque on a massless pulley.
Vigorous said:
the system comes to rest at two instants when the 8kg reaches maximum height and when it drops under free fall and hits the ground.
The 8kg cannot reach the ground again. Some mechanical energy must have been lost.
But the system will be instantaneously at rest again, perhaps many times.
Vigorous said:
K(h=0)-K(h=5)=0(T-14g)-5(T-14g)
I cannot figure out what you are doing there. Please clarify what each term represents.
 
  • #16
Change in kinetic energy of the mass when its about to hit the ground and 5 m off the ground=work done by the net force acting on the 14 kg mass.
Net force acting on the 14 kg mass=T-14g=-14a =-14(6g/22)+14g=>T=14g-42g/11. Thenet force is constant is not varying with height. (0-5)-displacement
So ΔK=K(height=0)-K(height=5)=(14g-42g/11-14g)(0-5)=-42g/11(0)+42g/11(5)
K(height=0)+42g/11(0)=K(height=5)+42g/11(5)
This combination of kinetic plus potential is conserved, until it hits the ground where this energy dissipates as heat or sound.
 
  • #17
Vigorous said:
Change in kinetic energy of the mass when its about to hit the ground and 5 m off the ground=work done by the net force acting on the 14 kg mass.
Net force acting on the 14 kg mass=T-14g=-14a =-14(6g/22)+14g=>T=14g-42g/11. Thenet force is constant is not varying with height. (0-5)-displacement
So ΔK=K(height=0)-K(height=5)=(14g-42g/11-14g)(0-5)=-42g/11(0)+42g/11(5)
K(height=0)+42g/11(0)=K(height=5)+42g/11(5)
This combination of kinetic plus potential is conserved, until it hits the ground where this energy dissipates as heat or sound.
Using energy you do not need to find the tension or acceleration.
I strongly recommend ignoring the given numbers and working entirely symbolically until the end. Write M for the larger mass, m for the smaller, h0 for the initial height.
The simplest way is:
Loss of PE =(M-m)g.h0.
Gain in KE = ½(M+m)v2.
##v^2=2gh_0\frac{M-m}{M+m}##.

We need to move on to the next phase. What happens after M hits the ground?
 
  • #18
haruspex said:
Using energy you do not need to find the tension or acceleration.
I strongly recommend ignoring the given numbers and working entirely symbolically until the end. Write M for the larger mass, m for the smaller, h0 for the initial height.
The simplest way is:
Loss of PE =(M-m)g.h0.
Gain in KE = ½(M+m)v2.
##v^2=2gh_0\frac{M-m}{M+m}##.
We need to move on to the next phase. What happens after M hits the ground?

For each mass, there is tension in the string, so there is work done by the tension in each mass. Energy conservation of mass M: The work done by the bigger mass M is positive because the block is moving in the direction of Mg. The Work done by Tension is negative because the block is moving counter to the direction of tension
1/2Mv2^2+Mg*h2-T*h2=1/2Mv1^2+Mgh1-Th1
h2=0 h1=5 v1=0
Why is this formulation of the conservation of energy for mass M incorrect?
 
  • #19
Vigorous said:
Why is this formulation of the conservation of energy for mass M incorrect?
I think you are taking the heavier mass as a system and the smaller mass as another system. But the recommended solution considers the two connected masses and pulley as a system, in which case the tension is an internal force and not an external force. Internal forces are to be ignored when applying conservation of energy law. So, tension is ignored in the recommended solution. Only external forces are to be considered when applying law of conservation of energy to a system.
 
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  • #20
haruspex said:
But the system will be instantaneously at rest again, perhaps many times.
When the lighter mass descends under gravity from its max height while the heavier mass is still sitting on the ground, then what would be the behavior of system as soon the string becomes taut? Will the heavier mass move up or the lighter mass will instantly comes to rest? This part would confuse most students.
 
  • #21
vcsharp2003 said:
When the lighter mass descends under gravity from its max height while the heavier mass is still sitting on the ground, then what would be the behavior of system as soon the string becomes taut? Will the heavier mass move up or the lighter mass will instantly comes to rest? This part would confuse most students.
Hence my post #4. To solve it readily you need to know you can treat it as a coalescence.
 
  • #22
haruspex said:
Hence my post #4. To solve it readily you need to know you can treat it as a coalescence.
But just out of curiosity, how would the masses behave when lighter mass during its descent makes the string taut?
 
  • #23
vcsharp2003 said:
But just out of curiosity, how would the masses behave when lighter mass during its descent makes the string taut?
AS I posted, it is like a coalescence. The two will move at the same speed.
Consider the impulsive tension in the string and apply conservation of momentum and angular momentum.
 
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  • #24
haruspex said:
AS I posted, it is like a coalescence. The two will move at the same speed.
Consider the impulsive tension in the string and apply conservation of momentum and angular momentum.

When system finally comes to rest according to part iii of the question, then it's difficult to know the position of masses, unless one assumes that lighter mass will finally come to rest so that the heavier mass is still sitting on ground and lighter mass is at rest at a height of 5 m, so that tension in string is equal to the weight of lighter mass.

Would above be a good assumption?
 
  • #25
vcsharp2003 said:
When system finally comes to rest according to part iii of the question, then it's difficult to know the position of masses, unless one assumes that lighter mass will finally come to rest so that the heavier mass is still sitting on ground and lighter mass is at rest at a height of 5 m, so that tension in string is equal to the weight of lighter mass.

Would above be a good assumption?
Yes, there cannot be any other stable state.
This could take a theoretically infinite number of stages, yet be completed in finite time.
 
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  • #26
haruspex said:
Yes, there cannot be any other stable state.
This could take a theoretically infinite number of stages, yet be completed in finite time.
It's not easy guessing the final rest position for a high school student.
 
  • #27
Would the 8-kg mass be a ble to lift the 14-kg mass when reaching 5 m of height on its way down?
 
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  • #28
Lnewqban said:
Would the 8-kg mass be a ble to lift the 14-kg mass when reaching 5 m of height on its way down?
The doubt here is whether 8 kg mass will instantly decelerate from some velocity to 0 m/s. Normally, some distance needs to be traveled to decelerate to 0 m/s.
 
  • #29
vcsharp2003 said:
It's not easy guessing the final rest position for a high school student.
Really? I would have thought many ten year olds could do it.
vcsharp2003 said:
The doubt here is whether 8 kg mass will instantly decelerate from some velocity to 0 m/s. Normally, some distance needs to be traveled to decelerate to 0 m/s.
Perhaps you are unfamiliar with impact problems. We accept that the impact takes some brief and unknown time, and the forces are large with unknown profile. But we can still deal with the momentum change, ΔP=∫F.dt.
 
  • #30
haruspex said:
Perhaps you are unfamiliar with impact problems. We accept that the impact takes some brief and unknown time, and the forces are large with unknown profile. But we can still deal with the momentum change, ΔP=∫F.dt.
The impulsive force in this case would be a variable tension force that is going to be very high in magnitude as compared to gravity or normal reaction forces and would act for a very short interval of time just after string becomes taut. This impulse would be on each of the masses. Is this idea of impulsive force correct?
 
  • #31
Lnewqban said:
Would the 8-kg mass be a ble to lift the 14-kg mass when reaching 5 m of height on its way down?
I think if there is a large impulsive force through tension that acts on heavier mass for a very small ##\triangle {t}## time interval then yes the heavier mass would get lifted off the ground when the string becomes suddenly taut.
 
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  • #32
vcsharp2003 said:
Only external forces are to be considered when applying law of conservation of energy to a system.
I got you! The work done by the tension of the rope which influence the motion of the masses m and M cancel each other. The string is inelastic, if M moves down by 5 m, m will move up by 5 m.
Energy conservation becomes:
Mgh0+mg(0)+1/2M(0)^2+1/2m(0)^2=Mgh0=Mg(0)+mgh0+1/2mv^2+1/2Mv^2 => gh0(M-m)=1/2v^2(M+m)
v=(2gh0(M-m)/(M+m))^1/2=5.169 m/s
The heavy mass hits the ground, its kinetic energy is transformed into another form of energy (heat or sound). The system will try to stay in motion at 5.169 m/s. The string connecting the two masses is no longer stretched because the weight of M is compensated by the reaction of the floor, M stays at rest. T no longer acts on m. m will uniformly decelerate with initial speed of 5.169 m/s
mgh0+1/2mv^2=mgh
h=6.36 m
 
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  • #33
vcsharp2003 said:
The impulsive force in this case would be a variable tension force that is going to be very high in magnitude as compared to gravity or normal reaction forces and would act for a very short interval of time just after string becomes taut. This impulse would be on each of the masses. Is this idea of impulsive force correct?
Yes.
 
  • #34
Vigorous said:
I got you! The work done by the tension of the rope which influence the motion of the masses m and M cancel each other. The string is inelastic, if M moves down by 5 m, m will move up by 5 m.
Energy conservation becomes:
Mgh0+mg(0)+1/2M(0)^2+1/2m(0)^2=Mgh0=Mg(0)+mgh0+1/2mv^2+1/2Mv^2 => gh0(M-m)=1/2v^2(M+m)
v=(2gh0(M-m)/(M+m))^1/2=5.169 m/s
The heavy mass hits the ground, its kinetic energy is transformed into another form of energy (heat or sound). The system will try to stay in motion at 5.169 m/s. The string connecting the two masses is no longer stretched because the weight of M is compensated by the reaction of the floor, M stays at rest. T no longer acts on m. m will uniformly decelerate with initial speed of 5.169 m/s
mgh0+1/2mv^2=mgh
h=6.36 m
For part iii, final rest position will be with lighter mass at 5 m position and heavier mass on ground. This final position will be reached after multiple up and down motions of lighter and heavier masses. The impulsive force on heavier mass will lift it for each such motion, but this impulsive force will lessen after each impact of heavier mass with the ground till it's no more there. You can see that each impact of heavier mass with ground will result in dissipation of energy. At final rest position you would be left with only PE of lighter mass i.e mgh as the final total energy of the system.
 

Related to What is the energy conservation principle in a two masses and pulley problem?

1. What is the energy conservation principle in a two masses and pulley problem?

The energy conservation principle states that the total energy in a closed system remains constant over time. In a two masses and pulley problem, this means that the total kinetic and potential energy of the system will remain the same, even as the masses and pulley move and transfer energy between each other.

2. How is energy conserved in a two masses and pulley problem?

In a two masses and pulley problem, energy is conserved through the transfer of potential and kinetic energy between the two masses and the pulley. As one mass moves down, it gains kinetic energy while the other mass moves up and loses kinetic energy. The potential energy of the system is also constantly changing as the masses move up and down.

3. Can energy be lost in a two masses and pulley problem?

No, energy cannot be lost in a two masses and pulley problem. The energy conservation principle states that the total energy in a closed system remains constant. While energy may be transferred between the masses and the pulley, the total energy of the system will always remain the same.

4. How does the energy conservation principle apply to real-life situations?

The energy conservation principle applies to all physical systems, including real-life situations. For example, in a simple pendulum, the total energy of the system (kinetic and potential) remains constant as the pendulum swings back and forth. This principle is also used in many engineering and scientific applications, such as in the design of energy-efficient machines and devices.

5. What are some potential sources of error when applying the energy conservation principle in a two masses and pulley problem?

Some potential sources of error when applying the energy conservation principle in a two masses and pulley problem include friction, air resistance, and measurement errors. Friction and air resistance can cause some energy to be lost to the surroundings, while measurement errors can lead to inaccurate calculations of the total energy in the system. It is important to minimize these sources of error in order to accurately apply the energy conservation principle.

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