Question
A 59 g particle is moving to the left at 29 m/s. How much net work must be done on the particle to cause it to move to the right at 38 m/s?
.....i have tried two different ways and gotten the wrong answer.
Kinetic Energy = 0.5 * Mass * (Change in Velocity)^2.
Change in Velocity here is 29+38=67m/s
Convert the Mass to kilograms: 0.059 kg
0.5 * 0.059 *67^2 = 132J
...then
Work is the change in kinetic energy. Since the particle of mass 64 (kilograms, pounds, whatever) is moving to the left at 27 (m/s, mph, whatever), you want to find how much work is require to completely change the particle's direction, and add velocity the other way. The change in velocity will NOT be 38-27, because that would simply be an increase of velocity in the direction it is current heading. Instead, the change in velocity will be 27 + 38, so 65. You have to overcome the velocity to the left, then add velocity to the right. Here's what you can do:
Work = KEf - KEo
KE = (1/2)mV^2
KEleft = (1/2)(64)(27)^2 = how much energy the particle has going left
KEstop = (1/2)(64)(0)^2 = how much energy when the particle is stopped
KEright = (1/2)(64)(38)^2 = how much energy the particle has going to the right
KEleft - KEstop = work to stop particle
KEstop - KEright = work to speed up the particle to the right (in the oppsite direction, but since energy is not a vector, you can remove the negative sign)
You can add the two numbers together, and you get your final answer
..i got 1.121 J
neither one of these answers are correct. Can anyone help me?
.....i have tried two different ways and gotten the wrong answer.
Kinetic Energy = 0.5 * Mass * (Change in Velocity)^2.
Change in Velocity here is 29+38=67m/s
Convert the Mass to kilograms: 0.059 kg
0.5 * 0.059 *67^2 = 132J
...then
Work is the change in kinetic energy. Since the particle of mass 64 (kilograms, pounds, whatever) is moving to the left at 27 (m/s, mph, whatever), you want to find how much work is require to completely change the particle's direction, and add velocity the other way. The change in velocity will NOT be 38-27, because that would simply be an increase of velocity in the direction it is current heading. Instead, the change in velocity will be 27 + 38, so 65. You have to overcome the velocity to the left, then add velocity to the right. Here's what you can do:
Work = KEf - KEo
KE = (1/2)mV^2
KEleft = (1/2)(64)(27)^2 = how much energy the particle has going left
KEstop = (1/2)(64)(0)^2 = how much energy when the particle is stopped
KEright = (1/2)(64)(38)^2 = how much energy the particle has going to the right
KEleft - KEstop = work to stop particle
KEstop - KEright = work to speed up the particle to the right (in the oppsite direction, but since energy is not a vector, you can remove the negative sign)
You can add the two numbers together, and you get your final answer
..i got 1.121 J
neither one of these answers are correct. Can anyone help me?
Answers
First, I do not know if the mass is 59 grams or 64 grams. You used each at different places. Computer exercises usually change the numbers when you make a second attempt. I don't know if that's what happened.
Secondly, you did not show your intermediate answers in the last two paragraphs, but they don't add up to 1.121 J like you said. I get 23.3J for stopping the particle (64 g), and about double that for pushing it to the right.
That makes a total of about 70 J.
Can you show your calculations?
Secondly, you did not show your intermediate answers in the last two paragraphs, but they don't add up to 1.121 J like you said. I get 23.3J for stopping the particle (64 g), and about double that for pushing it to the right.
That makes a total of about 70 J.
Can you show your calculations?
You would use the equation W=1/2(m)(v2^2-v1^2)
where m = mass, v2 = final velocity, and v1 is initial velocity
mass must be in kilograms
so it would be W=1/2(0.059)((38)^2-(29)^2)
and W=17.7885 J
where m = mass, v2 = final velocity, and v1 is initial velocity
mass must be in kilograms
so it would be W=1/2(0.059)((38)^2-(29)^2)
and W=17.7885 J
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