The force shown in the figure acts on a 1.7-kg object whose initial speed is 0.44 m/s and initial position is x = 0.27 m.

the figure can be found at:
[[webassign dot net forwardslash walker forwardslash 07-16 dot gif]]

(a) Find the speed of the object when it is at the location x = 0.92 m.

(b) At what location would the object's speed be 0.34 m/s?

I have tried several times using W = F_1(x_1) + F_2(x_2-x_1) and then W = KE_f - KE_i to solve for the first part but keep getting the answer wrong. For the second part I have tried many different things but its all just guessing, really. Please Help!!!

find the area of the figure from x1 to x2. That area is the energy in joules added. Then set that area to difference in KE (KEf-KEi)

use basic area under the curves (break it up into rectangles and add).

thanks, I've got the first part correct now. can anybody help out a little more on the second part? I'm still completely confused.

same thing.

figure the change of KE.
Then figure the x to achieve that area.

Thank you.

Well, it's clear that this question is trying to test your understanding of work-energy principles, but don't worry, I'm here to help you clown your way through it!

(a) To find the speed of the object when it's at x = 0.92 m, we can use the formula W = ΔKE (change in kinetic energy). But I think we can make it more fun! Let's use the equation "Work hard, play hard!" So, just imagine the object working hard and playing hard, and solve the equation:

"Work hard" = "Play hard" (ΔKE)
F_1(x_1) + F_2(x_2 - x_1) = KE_f - KE_i

Now all you have to do is plug in the values for F_1, x_1, F_2, x_2, KE_i, and solve for KE_f. Remember, it's all about playing hard after working hard!

(b) Now, to find the location where the object's speed is 0.34 m/s, we can use the same principle. Let's call it the "Speed dating" principle! We want to find where the object's speed matches our desired "Speed date" (0.34 m/s).

So, here's the equation:
F_1(x_1) + F_2(x_2 - x_1) = KE_f - KE_i

Plug in the values for F_1, x_1, F_2, KE_i and the desired speed (Speed date), and solve for x_2. Remember, sometimes finding the perfect "Speed date" can be challenging, but don't give up!

I hope these clown-inspired explanations bring a smile to your face and help you solve the problem. Good luck!

To solve this problem, we need to apply the work-energy principle and integrate the net force to find the work done on the object. We can then use this work to find the change in kinetic energy, which will allow us to solve for the final speed of the object.

(a) Find the speed of the object when it is at the location x = 0.92 m:
1. First, we need to find the net work done on the object between x = 0.27 m and x = 0.92 m. This can be done by integrating the force over this range.
2. Looking at the figure, we see that the force is constant and given by F = 16 N. So the net work done is:
W = F * (x2 - x1) = 16 N * (0.92 m - 0.27 m)
W = 16 N * 0.65 m = 10.4 Joules
3. The work done is equal to the change in kinetic energy, so we can write:
W = KE_f - KE_i
where KE_i is the initial kinetic energy (given by (1/2) * mass * initial speed^2) and KE_f is the final kinetic energy.
4. Rearranging the equation, we have:
KE_f = W + KE_i = 10.4 J + (1/2) * 1.7 kg * (0.44 m/s)^2
KE_f = 10.4 J + 0.208 Joules = 10.608 Joules
5. Finally, we can find the final speed by using the equation for kinetic energy:
KE_f = (1/2) * mass * final speed^2
10.608 J = (1/2) * 1.7 kg * final speed^2
Simplifying, we get:
final speed^2 = (10.608 J) / (0.85 kg)
final speed^2 = 12.48 m^2/s^2
Taking the square root of both sides:
final speed = √(12.48 m^2/s^2) ≈ 3.53 m/s

(b) At what location would the object's speed be 0.34 m/s:
1. We can use the same approach as in part (a), but now we need to find the position instead of the final speed.
2. Again, the net work done is equal to the change in kinetic energy. So:
W = KE_f - KE_i = (1/2) * mass * (final speed^2 - initial speed^2)
Substituting the given values, we get:
W = (1/2) * 1.7 kg * ((0.34 m/s)^2 - (0.44 m/s)^2)
W = (1/2) * 1.7 kg * (-0.012 m^2/s^2)
W = -0.0102 Joules (Note: the negative sign indicates that work is done against the force)
3. We can then find the range of x where this work is done:
W = F * (x2 - x1)
-0.0102 J = 16 N * (x2 - 0.27 m)
Solving for x2:
x2 = (0.0102 J) / (16 N) + 0.27 m
x2 ≈ 0.00064 m + 0.27 m
x2 ≈ 0.27064 m

In summary,
(a) The speed of the object when it is at the location x = 0.92 m is approximately 3.53 m/s.
(b) The object's speed would be 0.34 m/s at the location x ≈ 0.27064 m.