Kinetic Energy and Speed: Doubling KE with Multiplying Speed?

Click For Summary
SUMMARY

The discussion centers on the relationship between kinetic energy (KE) and speed (v) in physics, specifically how to double the kinetic energy of a mass. The key equation is KE = 1/2 mv². To double the kinetic energy, the speed must be multiplied by the square root of 2, resulting in a factor of approximately 1.41, not 4 as initially suggested. The algebraic approach confirms that if KE_f = 2 KE_i, then v_f = v_i * √2.

PREREQUISITES
  • Understanding of kinetic energy formula: KE = 1/2 mv²
  • Basic algebra skills for manipulating equations
  • Familiarity with square roots and their properties
  • Concept of proportional relationships in physics
NEXT STEPS
  • Study the derivation of kinetic energy equations in classical mechanics
  • Learn about the implications of energy conservation in physics
  • Explore the relationship between potential energy and kinetic energy
  • Investigate real-world applications of kinetic energy in engineering
USEFUL FOR

Students studying physics, educators teaching mechanics, and anyone interested in understanding the principles of energy and motion.

majormuss
Messages
124
Reaction score
4

Homework Statement


If the Kinetic energy of a given mass is to be doubled, its speed must be multiplied by?


Homework Equations



KE= 1/2mv^2

The Attempt at a Solution


I believe the answer is 4, but am not really sure. I figured that if the Kinetic energy should be doubled, then since the v(velocity) on the other hand is squared and the proportionality also seems to be direct then the answer is 2^2=4. Am I right?
 
Physics news on Phys.org
No, that is not right. For problems like this it is worth picking some numbers to substitute in for variables. As a hint, you aren't going to get a whole number.
 
KE=.5*m*V1^2 therefore V1=Sqrt(2*KE/m)

2*KE=.5^V2^2 therefore V2=sqrt(4*KE/m)

What must you multiply V1 by to get V2?

V1*Z=V2

Z=V1/V2= 2/sqrt(2)
 
u can think about it this way ... u have your KE(1) = 1/2mv^2 , so in order to have 2KE(1) which is (2)1/mv^2 .. what u should have with the velocity so that u get this factor of (2) ..
 
RTW69 said:
KE=.5*m*V1^2 therefore V1=Sqrt(2*KE/m)

2*KE=.5^V2^2 therefore V2=sqrt(4*KE/m)

What must you multiply V1 by to get V2?

V1*Z=V2

Z=V1/V2= 2/sqrt(2)

I think u have made some mistake here..
 
thebigstar25 said:
u can think about it this way ... u have your KE(1) = 1/2mv^2 , so in order to have 2KE(1) which is (2)1/mv^2 .. what u should have with the velocity so that u get this factor of (2) ..

still confused... is the answer 1/2?, which is it and how can arrive at the right answer.
 
Consider this: you have two kinetic energies, KEi (initial) and KEf (final). These kinetic energies are computed with the same mass, but with different velocities, vi and vf. You're given that KEf is twice KEi, and you're trying to figure out what the relationship between vi and vf is.

What you've been doing so far is kind of like a guess-and-check method: you start with
KE_i = \frac{1}{2}mv_i^2
KE_f = \frac{1}{2}mv_f^2
and try plugging in first v_f = 4v_i (your initial guess)
KE_f = \frac{1}{2}m(4 v_i)^2 = 16*\frac{1}{2}m v_i^2 = 16KE_i
...nope, that's not it. Now what if you try 1/2?
KE_f = \frac{1}{2}m\biggl(\frac{1}{2} v_i\biggr)^2 = \frac{1}{4}*\frac{1}{2}m v_i^2 = \frac{1}{4}KE_i
...that's not it either.

Try the algebraic approach. Start with the condition that you need to be true, that the final KE is twice the initial KE
KE_f = 2 KE_i
and substitute in the expressions for kinetic energy
KE_i = \frac{1}{2}mv_i^2
KE_f = \frac{1}{2}mv_f^2
Then cancel out the common factors on both sides of the equation and see what you're left with. Finally, what mathematical step should you take to get it into the form
v_i = \bigl(\cdots\bigr)v_f
and what is the coefficient in the parentheses? That will be the answer you're looking for.
 


diazona said:
Consider this: you have two kinetic energies, KEi (initial) and KEf (final). These kinetic energies are computed with the same mass, but with different velocities, vi and vf. You're given that KEf is twice KEi, and you're trying to figure out what the relationship between vi and vf is.

What you've been doing so far is kind of like a guess-and-check method: you start with
KE_i = \frac{1}{2}mv_i^2
KE_f = \frac{1}{2}mv_f^2
and try plugging in first v_f = 4v_i (your initial guess)
KE_f = \frac{1}{2}m(4 v_i)^2 = 16*\frac{1}{2}m v_i^2 = 16KE_i
...nope, that's not it. Now what if you try 1/2?
KE_f = \frac{1}{2}m\biggl(\frac{1}{2} v_i\biggr)^2 = \frac{1}{4}*\frac{1}{2}m v_i^2 = \frac{1}{4}KE_i
...that's not it either.


Try the algebraic approach. Start with the condition that you need to be true, that the final KE is twice the initial KE
KE_f = 2 KE_i
and substitute in the expressions for kinetic energy
KE_i = \frac{1}{2}mv_i^2
KE_f = \frac{1}{2}mv_f^2
Then cancel out the common factors on both sides of the equation and see what you're left with. Finally, what mathematical step should you take to get it into the form
v_i = \bigl(\cdots\bigr)v_f
and what is the coefficient in the parentheses? That will be the answer you're looking for.
Thanks a lot for the hint... really appreciate it. so my answer turns out 2^1/2
I figured it logically: since 'v' is squared, what number when squared gives 2.
In other words 1.41
 
2/sqrt(2) is 1.41 as described above
 

Similar threads

  • · Replies 21 ·
Replies
21
Views
5K
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
992
  • · Replies 12 ·
Replies
12
Views
2K
Replies
5
Views
934
  • · Replies 8 ·
Replies
8
Views
1K
  • · Replies 13 ·
Replies
13
Views
1K
Replies
1
Views
2K
Replies
2
Views
1K
  • · Replies 33 ·
2
Replies
33
Views
3K