Figure can be seen here: https://session.masteringphysics.com/problemAsset/…
Part A:
Let the zero of the potential energy be at x = 0 m.What is the potential energy at x = 1.0, 2.0, 3.0, and 4.0 m? Hint: Think about the definition of potential energy and the
geometric interpretation of the work done by a varying force.
Express your answers using two significant figures separated by commas.
Part B:
Suppose the particle is shot toward the right from x = 1.0 m with a speed of 19 m/s. Where is the particle’s turning point?
1 Answer
-
(a) Due to the relationship W=F∆d, as well as W=∆U+∆K, we know that in this case U=F∆x.
If you take the area under the curve for each x-value, you ll get the U (aka PE).
20,40,60,70 is your answer
(b) For this part, use the formula for Kinetic Energy, K=1/2mv^2 to find K at the given speed. K=1/2(0.1)(19^2)= 18.05 N.
Now, using the W=F∆d formula again, we can say K=F∆d, or K=F(x2-x1).
Using the plot from above, we can infer that the force is still 20N.
Since we start at x=1.0m, x1=1.
Solving foe x2, we get x2=(K/F)+x1
So, x2 =(18.05/20)+1=1.9m
The turning point is 1.9m.
Source(s): Was forced to complete the damn Mastering Physics problem on my own since I couldn t find the answer anywhere else.
