Can the force of 1.50N accelerate a 0.20kg cart to a speed of 6.70m/s?

In summary: Oh really thank you so much for clearing my doubt. This was the only thing that I wanted to know. Thank you so so much!In summary, the conversation discusses a question regarding the use of either work energy theorem or Newton's law in solving a numerical problem. The expert explains that the two methods should give the same result as the work-energy theorem is derived from Newton's law. However, the conversation also highlights a mistake in the calculation of acceleration using Newton's law, leading to a wrong answer. The expert clarifies that the correct value of acceleration should be 7.5 m/s^2, and using this value, both methods will yield the same result. Overall, the conversation emphasizes the importance of correctly applying the laws and
  • #1
shayaan_musta
209
2
Work energy or Newton's law?

I have a question and solution(from book) too:
Question:
A force of 1.50N acts on a 0.20kg cart so as to accelerate it along an air track. The track and force are horizontal and in line. How fast is the cart going after acceleration from rest through 30cm, if friction is negligible?

Answer:
The work done by the force causes, and is equal to, the increase in K.E. of the cart, Therefore,
Work done=(K.E.)end - (K.E.)start
or Fs cos0°=½mvf2 - 0
substituting gives
(1.50N)(0.30m)=½(0.20kg)vf2
from which vf=2.1m/s.


Now my problem is:
Can't this problem be solve with the help of equations of motion?
means, we can do this as,
F=ma
a=15/0.2
a=75m/s²
using 3rd equation of motion,
2as= vf2 - vi2
putting "a" in above equation,
2(75)0.3=vf2 - 0
vf2=45
vf= 6.70m/s

But my approach is wrong and the book solution is correct. Can anyone tell me that how we differentiate between numerical either it is to be solved by work energy theorem or by Newton's law? Plzzzz
 
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  • #2


shayaan_musta said:
I have a question and solution(from book) too:
Question:
A force of 1.50N acts on a 0.20kg cart so as to accelerate it along an air track. The track and force are horizontal and in line. How fast is the cart going after acceleration from rest through 30cm, if friction is negligible?

Answer:
The work done by the force causes, and is equal to, the increase in K.E. of the cart, Therefore,
Work done=(K.E.)end - (K.E.)start
or Fs cos0°=½mvf2 - 0
substituting gives
(1.50N)(0.30m)=½(0.20kg)vf2
from which vf=2.1m/s.

Now my problem is:
Can't this problem be solve with the help of equations of motion?
means, we can do this as,
F=ma
a=15/0.2
a=75m/s²
using 3rd equation of motion,
2as= vf2 - vi2
putting "a" in above equation,
2(75)0.3=vf2 - 0
vf2=45
vf= 6.70m/s

But my approach is wrong and the book solution is correct. Can anyone tell me that how we differentiate between numerical either it is to be solved by work energy theorem or by Newton's law? Plzzzz

They should give the same result.
Multiply the 3rd equation of motion by mass & divide by 2, then use F=ma: you get the Work - Energy equation you used in the first place. They have to give the same result!
 
Last edited:
  • #3


The acceleration you calculated is a bit high. An order of magnitude too high. Should be 7.5 m/s2.
 
  • #4


SammyS said:
They should give the same result.
Multiply the 3rd equation of motion by mass & divide by 2, then use F=ma: you get the Work - Energy equation you used in the first place. They have to give the same result!

No no no no.. Wait wait..
I am trying to ask you to differentiate between both type of numerical?
Let, if I didn't write any title "Work energy or Newton's law?" then which method could you use, Work energy Theorem or Newton's law?
 
  • #5


gneill said:
The acceleration you calculated is a bit high. An order of magnitude too high. Should be 7.5 m/s2.

Ok you are saying that my calculated acceleration is of high magnitude therefore it can't be solved by Newton's law(or by my method). Is it??
 
  • #6


You plugged in 15 N for force (instead of the given 1.5 N) when applying Newton's second law, so you got a wrong acceleration value.

a = F/m = 1.5 /0.2 = 7.5 m/s^2 instead of 75.

vf2=2as=2*7.5*0.3 = 4.5---> vf=2.1 m/s

The work-energy theorem is derived from Newton's law. They are equivalent.

ehild
 
  • #7


ehild said:
You plugged in 15 N for force (instead of the given 1.5 N) when applying Newton's second law, so you got a wrong acceleration value.

a = F/m = 1.5 /0.2 = 7.5 m/s^2 instead of 75.

vf2=2as=2*7.5*0.3 = 4.5---> vf=2.1 m/s

The work-energy theorem is derived from Newton's law. They are equivalent.

ehild

Oh thanks thanks. I see. It means that from any method I would get same answer.
 

FAQ: Can the force of 1.50N accelerate a 0.20kg cart to a speed of 6.70m/s?

What is the relationship between work, energy, and Newton's laws?

The concept of work and energy is closely related to Newton's laws of motion. According to the first law, an object in motion will remain in motion unless acted upon by a force. This means that in order for an object to change its motion (i.e. speed up, slow down, change direction), a force must be applied to it. Work is defined as the product of force and displacement, and it is through the application of work that energy is transferred to or from an object, causing it to change its motion in accordance with Newton's laws.

What is the difference between potential and kinetic energy?

Potential energy is the stored energy an object possesses due to its position or configuration. It is often described as the "energy of position" and depends on factors such as the object's height, mass, and the strength of the gravitational field. On the other hand, kinetic energy is the energy an object possesses due to its motion. It is described as the "energy of motion" and depends on the mass and velocity of the object. The total energy of an object is the sum of its potential and kinetic energy.

How does work-energy theorem relate to Newton's second law?

The work-energy theorem states that the net work done on an object is equal to the change in its kinetic energy. This means that the work done by all the forces acting on an object will result in a change in the object's kinetic energy, in accordance with Newton's second law (force equals mass times acceleration). Essentially, the work-energy theorem is a mathematical representation of Newton's second law in terms of energy.

Can you give an example of how to calculate work and energy in a real-world scenario?

One example is calculating the work and energy involved in lifting a book from the ground to a shelf. If the book has a mass of 1kg and the shelf is 2 meters high, the work done to lift the book would be equal to the product of the force exerted (9.8N due to gravity) and the distance moved (2m), which is 19.6 joules. This work done transfers potential energy to the book, giving it a new potential energy of 19.6 joules. As the book falls, this potential energy is converted into kinetic energy, and the book reaches a speed of √(2 x 9.8 x 2) = 6.26 m/s just before it reaches the ground.

Can the law of conservation of energy be applied to work and Newton's laws?

Yes, the law of conservation of energy states that energy cannot be created or destroyed, only transferred or converted from one form to another. This applies to work and Newton's laws in the sense that the work done on an object will result in a change in its energy, but the total amount of energy in the system will remain constant. This means that the energy transferred through work is equal to the change in energy of the object, and this energy can be converted to different forms, such as potential or kinetic energy, in accordance with Newton's laws.

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