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

[shayaan_musta]

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

 

[SammyS]

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!

 

[gneill]

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

 

[shayaan_musta]

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?

 

[shayaan_musta]

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??

 

[ehild]

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

 

[shayaan_musta]

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.