<p>I'm have trouble with how to solve this question, and I dont know why because it seems like an easy question:</p>
<p>An object released from rest at time t=0 slides down a frictionless incline a distance of 1 meter during the first second. The distance traveled by the object during the time interval from t=1 second to t=2 seconds is
A. 1 m
B. 2 m
C. 3 m
D. 4 m
E. 5 m</p>
<p>the correct answer is C</p>
<p>The change in velocity during the first second has to be 2 m/s b/c the object travels 1 meter during 1 second. So the acceleration is 2 m/s^2. So you know the v=2 at t=1 and v=4 at t=2. So it travels 3 meters during that time.
Sorry if that sounds confusing, there's probably a simple kinematic for it, but that's how i would do it.</p>
<p>i dont know how the change in velocity is 2 m/s because if the object moved 1 m in 1 sec, wouldnt it have a velocity of 1 m/s???</p>
<p>This is from the 2004 test:
How much net work must be done by an external force to move a -1 uC point charge from rest at point C ( V= 10V) to rest at point E (V = 20V).</p>
<p>10 uJ or -10 uJ ? Why?</p>
<p>Also, when given a diagram of equipotential lines, how do you tell at which point the electric field has the greatest potential?</p>
<p>eternity1115 - the average velocity*time=distance if acceleration is uniform
since time =1 and distance = 1, average velocity must be 1.
average velocity is 1/2(vmax-vmin) since vmin = 0, vmax must be 2.</p>
<p>You have the 2004 test?! could you please send it to me? <a href="mailto:kishenr@gmail.com">kishenr@gmail.com</a></p>
<p>me too please: <a href="mailto:amit.talapatra@gmail.com">amit.talapatra@gmail.com</a></p>
<p>Same here..Plz
<a href="mailto:iyolinjc@hanmail.net">iyolinjc@hanmail.net</a></p>
<p>use the equation deltaX = v0t + 1/2at^2, u should get 3, but my question is that since it is an incline, shouldnt the acceleration be gsin(theta)?
Thanks...</p>
<p>yeah, but since theta isnt changing the acceleration is constant anyway (assume constant g when near the earth's surface)</p>
<p>ok this is the way I did it.
F=mgsin(theta) since its frictionless
ma=mgsin(theta) m's cancel
a=gsin(theta) acceleration is a constant number here
v=gt sin(theta) v0 is 0 because its starts from rest
x=(g/2)sin(theta) *(t^2) (dont' worry about +C because we are finding difference between x(2)-x(1) which makes the C's canel out anyways).
from what we know...1= (g/2)sin(theta)=5sin(theta)
for x(2)=(g/2)sin(theta) *(4) = 20 sin(theta)
x(2)-x(1)=15sin(theta)
remember x(1)=5sin(theta)=1
so 15 sin(theta) must equal 3</p>
<p>Hope that's clear enough</p>
<p>When I did that question I did it like this...</p>
<p>delta x = 1/2 at^2 + 0 [Vo = 0]
1 = (1/2)(a)(1)^2
a = 2</p>
<p>delta x = 1/2 at^2
delta x = (1/2)(2)(2)^2
delta x = 4</p>
<p>so it traveled 4 meters in 2 seconds
since it traveled 1 meter in 1 second, it traveled the last 3 from t=1 to t=2</p>
<p>physics08, since nobody bothered to answer your question:</p>
<p>its -10 uJ. to move a positive point charge from lower to higher potential requires positive work. but since you have a negative charge, you need negative work. the rest is easy: 1uC * 10 V = 10 uJ</p>