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?

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

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?

Sure! Let me try to help you out with some clown humor.

Well, it seems like calculating work is no laughing matter! But don't worry, I've got a silly solution for you.

Instead of doing all those complicated calculations, let's try a different approach. Think of the particle as a comically lazy cartoon character. To get them to move in the opposite direction, we need to motivate them with a lot of work!

So, how much work do we need to do? Well, imagine if you had to move a super heavy clown nose. You'd have to put in a lot of effort, right? The same idea applies here.

Since the particle is already moving to the left, we need to counteract that motion and then add velocity to the right. So, we need to do double the work!

Let's say the work to stop the particle from moving to the left is W1, and the work to speed it up to the right is W2. Then the total work we need is W1 + W2.

But since we need to do double the work, we can simply multiply the incorrect answer you got, 1.121 J, by 2! So, the correct answer should be 2.242 J.

I hope that puts a clownish smile on your face! Just remember, sometimes it takes a little extra silliness to find the right answer. Keep up the good work!

To find the net work done on the particle, you can use the work-energy principle. The work done on an object is equal to the change in its kinetic energy. The equation for work is:

Work = Change in Kinetic Energy

The third equation you mentioned for calculating the work is correct:

Work = KEf - KEo

where KEf is the final kinetic energy and KEo is the initial kinetic energy.

Let's calculate the net work step-by-step:

1. Calculate the initial kinetic energy (KEo):

KEo = 0.5 * mass * (initial velocity)^2
= 0.5 * 0.059 kg * (-29 m/s)^2 [since the particle is moving to the left]
= 24.879 J

2. Calculate the final kinetic energy (KEf):

Given that the particle is moving to the right at 38 m/s, the velocity would be positive:

KEf = 0.5 * mass * (final velocity)^2
= 0.5 * 0.059 kg * (38 m/s)^2
= 43.461 J

3. Calculate the net work:

Work = KEf - KEo
= 43.461 J - 24.879 J
= 18.582 J

Therefore, the net work required to cause the particle to move to the right at 38 m/s is 18.582 J.

To solve this problem correctly, we need to consider the work done on the particle in two separate steps: first, the work to bring the particle to rest, and second, the work to accelerate it in the opposite direction.

Step 1: Work to bring the particle to rest
The initial kinetic energy (KE) of the particle is given by the formula KE = (1/2) * mass * velocity^2. Plugging in the values, we find KE_left = (1/2)(0.059 kg)(29 m/s)^2 = 24.901 J.

Next, since the particle needs to be stopped, its final kinetic energy will be zero (KE_right = 0). Therefore, the work done to bring the particle to rest is given by Work_1 = KE_left - KE_right = 24.901 J - 0 J = 24.901 J.

Step 2: Work to accelerate the particle in the opposite direction
In this step, the initial velocity is zero (since the particle is at rest), and the final velocity is 38 m/s to the opposite direction. Therefore, the final kinetic energy is KE_right = (1/2)(0.059 kg)(38 m/s)^2 = 43.4826 J.

The work done to accelerate the particle in the opposite direction is given by Work_2 = KE_right - KE_left = 43.4826 J - 0 J = 43.4826 J.

Finally, the net work done on the particle is the sum of the work in step 1 and step 2:
Net Work = Work_1 + Work_2 = 24.901 J + 43.4826 J = 68.3836 J.

Therefore, the correct answer is 68.3836 J.