Please help me with this physics problem! I've tried using all possible resources given by the website and cannot figure it out. PRACTICE ITUse the worked example above to help you solve this problem. A proton is released from rest atx = -2.00 cm in a constant electric field with magnitude 1.50 x 103 N/C pointing in the positivex-direction.(a) Assuming an initial speed of zero, find the speed of a proton at x = 0.0300 m with a potentialenergy of –1.20 × 10-17 J. (Assume the potential energy at the point of release is zero.)1.2e5m/s(b) An electron is now fired in the same direction from the same position. Find the initial speed ofthe electron (at x = -2.00 cm) given that its speed has fallen by half when it reachesx = 0.180 m, a change in potential energy of 4.81 x 10-17 J.1.19e7v m/sEXERCISEHINTS: GETTING STARTED I I'M STUCK!The electron in part (b) travels from x = 0.180 m (where it has half the initial speed you previouslycalculated) to x = -0.300 m within the constant electric field. If there's a change in electric potentialenergy of -1.15 × 10-16 J as it goes from x = 0.180 m to x = -0.300 m, find the electron's speed atx = -0.300 m. (Note: Use the values from the Practice It section. Account for the fact that the electronmay turn around during its travel.)Vxm/s

Name: Please help me with this physics problem! I've tried using all possibl
Uploaded: 2024-03-05T11:56:34.000Z
Channel: Bartleby
Description: Please help me with this physics problem! I've tried using all possible resources given by the website and cannot figure it out. PRACTICE ITUse the worked example above to help you solve this problem. A proton is released from rest atx = -2.00 cm in

Answered: Use the worked example above to help…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

See similar textbooks

Related questions

Concept explainers

Topic Video

Question

Please help me with this physics problem! I've tried using all possible resources given by the website and cannot figure it out.

PRACTICE IT
Use the worked example above to help you solve this problem. A proton is released from rest at
x = -2.00 cm in a constant electric field with magnitude 1.50 x 103 N/C pointing in the positive
x-direction.
(a) Assuming an initial speed of zero, find the speed of a proton at x = 0.0300 m with a potential
energy of –1.20 × 10-17 J. (Assume the potential energy at the point of release is zero.)
1.2e5
m/s
(b) An electron is now fired in the same direction from the same position. Find the initial speed of
the electron (at x = -2.00 cm) given that its speed has fallen by half when it reaches
x = 0.180 m, a change in potential energy of 4.81 x 10-17 J.
1.19e7
v m/s
EXERCISE
HINTS: GETTING STARTED I I'M STUCK!
The electron in part (b) travels from x = 0.180 m (where it has half the initial speed you previously
calculated) to x = -0.300 m within the constant electric field. If there's a change in electric potential
energy of -1.15 × 10-16 J as it goes from x = 0.180 m to x = -0.300 m, find the electron's speed at
x = -0.300 m. (Note: Use the values from the Practice It section. Account for the fact that the electron
may turn around during its travel.)
Vx
m/s

Expert Solution

Step 1 a) speed of proton

Electric field is conservative force field hence, there are total mechanical energy of system remains constant.

Hence, For the given case

$∆ V = - ∆ K w h e r e, ∆ V = c h a n g e i n p o t e n t i a l e n e r g y ∆ K = c h a n g e i n k i n e t i c e n e r g y V_{f} - V_{i} = K_{i} - K_{f} I n i t i a l k i n e t i c e n e r g y o f p r o t o n i s z e r o a n d I n i t i a l p o t e n t i a l e n e r g y o f p r o t o n a l s o z e r o . V_{f} = - K_{f} V_{f} = - \frac{1}{2} m v_{f}^{2} 1)$

Values given in the question are as follow:

$V_{f} = - 1.20 \times 10^{- 17} J m = 1.67 \times 10^{- 27} k g$

Substitute known values in equation-1):

$- 1.20 \times 10^{- 17} J = - \frac{1}{2} \times 1.67 \times 10^{- 27} k g \times v_{f}^{2} v_{f}^{2} = \frac{1.20 \times 10^{- 17} \times 2}{1.67 \times 10^{- 27}} {(\frac{m}{s})}^{2} v_{f}^{2} = 1.44 \times 10^{10} {(\frac{m}{s})}^{2} v_{f} = 1.2 \times 10^{5} \frac{m}{s}$

Speed of proton at distance x = 0.0300 m is $v_{f} = 1.2 \times 10^{5} \frac{m}{s}$

Trending now

This is a popular solution!

Step by step

Solved in 2 steps

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.

Similar questions