JUST NEED THE LAST PART! PRACTICE IT Use the worked example above to help you so

Question

JUST NEED THE LAST PART!

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 1.50 103 N/C pointing in the positive x-direction.

(a) Assuming an initial speed of zero, find the speed of a proton at x = 0.0500 m with a potential energy of -1.65 10-17 J.

1.42e5 m/s correct

(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) given that its speed has fallen by half when it reaches x = 0.120 m, a change in potential energy of 3.45 10-17 J.

9.92e6 m/s correct

EXERCISE

The electron in part (b) travels from x = 0.120 m to x = -0.180 m within the constant electric field. If there's a change in electric potential energy of -8.10 10-17 J as it goes from x = 0.120 m to x = -0.180 m, find the electron's speed at x = -0.180 m. (Use the values from the Practice It section. Account for the fact that the electron may turn around during its travel.)

???????????? = vx

Explanation / Answer

from (b), the speed at x1 = 0.120 m is v1 = 9.92e6/2 = 4.96e6 m/s
find speed v2 at x2 = -0.180 m
given potential energy U: U2 - U1 = -8.0e-17 J
since K1 + U1 = K2 + U2
m*v12/2 + U1 + m*v22/2 + U2
v2 = [v12 - 2(U2 - U1)/m] = 1.42e7 m/s

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