Modern Physics For Scientists And Engineers
2nd Edition
ISBN: 9781938787751
Author: Taylor, John R. (john Robert), Zafiratos, Chris D., Dubson, Michael Andrew
Publisher: University Science Books,
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Chapter 2, Problem 2.52P
To determine
An expression for the speed of a particle of mass m and charge q released from the origin in a uniform electric field E, as a function of the distance x and a graph between u and x.
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Since the initial potential energy and final kinetic energy are zero, our equation now becomes
EP,f = EK,i.
We can then substitute the formula for kinetic energy,
EK = 1/2mv2,
and the formula for gravitational potential energy,
EP = mgh,
mghf = 1/2mvi2.
Now it's just a matter of doing the algebra, solving for the final height
hf,
and substituting values to find
hf.
Notice that the mass m divides out of both sides of the equation, so the value of the mass is not needed to find the final height.
Calculate the maximum height of the ball in meters.
hf = __________ m
Go back to question 6 but this time assume uk=0.2.
a) How much time elapses before the block reaches its maximum height up the plane?
b) How much time elapses from the point it reaches maximum height up the plaane to the point where it was launched?
Consider the equation for kinetic energy: KE = 1/2mv^2 = 1/2 * m * v^2. If I ask you to take the derivative of kinetic energy, you should ask "the derivative with respect to what?"
a) Suppose mass m is constant. Compute the derivative of KE with respect to v, (d(KE)/dv).
b) Who takes derivatives with respect to velocity? No one. Except you, just now. Sorry.
The rate of change of energy with respect to time is more important: it is the Power. Now, consider velocity v to be a function of time, v(t). We will rewrite KE showing this time dependance: KE= 1/2 * m * v(t)^2. Show that (d(KE)/dt) = F(t)v(t). Hint: use Newton's second law, F = ma, to simplify.
c) In the computation above, we assumed m was constant, and v was changing in time. Think of a physical situation in which both m and v are varying in time.
d) Compute the Power when both mass and velocity are changing in time. (First rewrite KE(t) showing time dependence, then compute (d(KE)/dt).
Chapter 2 Solutions
Modern Physics For Scientists And Engineers
Ch. 2 - Prob. 2.1PCh. 2 - Prob. 2.2PCh. 2 - Prob. 2.3PCh. 2 - Prob. 2.4PCh. 2 - Prob. 2.5PCh. 2 - Prob. 2.6PCh. 2 - Prob. 2.7PCh. 2 - Prob. 2.8PCh. 2 - Prob. 2.9PCh. 2 - Prob. 2.10P
Ch. 2 - Prob. 2.11PCh. 2 - Prob. 2.12PCh. 2 - Prob. 2.13PCh. 2 - Prob. 2.14PCh. 2 - Prob. 2.15PCh. 2 - Prob. 2.16PCh. 2 - Prob. 2.17PCh. 2 - Prob. 2.18PCh. 2 - Prob. 2.19PCh. 2 - Prob. 2.20PCh. 2 - Prob. 2.21PCh. 2 - Prob. 2.22PCh. 2 - Prob. 2.23PCh. 2 - Prob. 2.24PCh. 2 - Prob. 2.25PCh. 2 - Prob. 2.26PCh. 2 - Prob. 2.27PCh. 2 - Prob. 2.28PCh. 2 - Prob. 2.29PCh. 2 - Prob. 2.30PCh. 2 - Prob. 2.31PCh. 2 - Prob. 2.32PCh. 2 - Prob. 2.33PCh. 2 - Prob. 2.34PCh. 2 - Prob. 2.35PCh. 2 - Prob. 2.36PCh. 2 - Prob. 2.37PCh. 2 - Prob. 2.38PCh. 2 - Prob. 2.39PCh. 2 - Prob. 2.40PCh. 2 - Prob. 2.41PCh. 2 - Prob. 2.42PCh. 2 - Prob. 2.43PCh. 2 - Prob. 2.44PCh. 2 - Prob. 2.45PCh. 2 - Prob. 2.46PCh. 2 - Prob. 2.47PCh. 2 - Prob. 2.48PCh. 2 - Prob. 2.49PCh. 2 - Prob. 2.50PCh. 2 - Prob. 2.51PCh. 2 - Prob. 2.52P
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