Ch6_64and72

a)W is defined by F`ds so W=-k/x^2 (x2-x1). Since k is negative and the `ds will be positive W should be negative for this case.

The force is variable. There is no single force on the interval, because x varies over the interval.

Can you set up a Riemann sum to represent the approximate work?

Start by finding the work done over a small interval of unspecified width `dx, and let x_rep be a representative value of x within the interval (since the interval is small, the representative point can be anywhere in the interval, so x_rep can be any x within the interval).

Then assume a series of small intervals `dx_1, `dx_2, ..., `dx_n with representative points x_rep_1, x_rep_2, ..., x_rep_n. Write the expression for `dW_i, where i is any number between 1 and n.

Then write the expression for sum(`dW_i), where the summation is assumed to go from i = 1 to n. Assuming that the intervals `dx_1, ..., `dx_n represent a partition of the interval from x = x1 to x = x2 (a partition is a subdivision of an interval into distinct subintervals), this expression will represent the total work done between x1 and x2.

If the 'mesh' of the partition then approaches 0 (the 'mesh' is the size of the largest of the subintervals), then n will approach infinity, and the sum will approach an integral. What is this integral?

Get back to me with the best answers you can provide to these questions, and of course with additional questions if they arise.

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Would the Riemann sum for the `dw be ∑-kx^2(x2-x1)? This just represents adding sectors of the equation together to get the area under the F vs x curve. This sum represents the integral, but the integral given in the query is k / x2 - k / x1. I am not sure how this was calculated, I know the integral is the 1/1+n t ^n+1.

On an interval of length `dx_i containing point x_rep_i the representative force would be -k / (x_rep_i)^2, the force would be applied thru distance `dx_i so the work would be `dW_i = (-k / (x_rep_i)^2) * `dx_i. The Riemann sum would be

sum(`dW_i, i from 1 to n) = sum(-k / (x_rep_i)^2) * `dx_i, i from 1 to n).

In the limit this sum approaches the integral int(-k / x^2 dx, x from x1 to x2.

An antiderivative of k / x^2 is - k / x (use your formula for n = -2, and note that the n in your formula has nothing to do with the n in the Riemann sum), so an antiderivative of -k / x^2 is k / x. Evaluating this between the limits x = x1 and x = x2 you get k / x1 - k/x2, which coincides with the solution in the text.

72) A net force with magnitude (5.00N/m^2)x^2 and directed at a constant angle of 31 degrees with the positive x axis acts on an object of mass .25 kg as the object moves parallel to the x axis. How fast is the object moving at x = 1.5 mif it has a speed of 4 m/s at x=1m? We know that F=ma and if the net force is 5.00x^2 we can use a={5.00x^2(cos31)}/.25, then we can use KE=F`ds, so that .5(.25)v^2=.25(17.14*1.5^2)(.5). Solving for v we get 6.21 m/s.

You have the right idea, but KE is not equal to F `ds. It is so that

`dKE = F_net `ds.

But `dKE is the change in KE, equal to KE_f - KE_0, the difference between initial and final KE.

Can you modify your solution accordingly?

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Ok so .5(.25)v^2-(.5)(.25)(4^2)=.25(17.14*1.5^2)(.5). Solving for v we get 14.1 m/s.

Good. Now you need to take account of the fact that the force is variable. To find the work done by the force, you're going to have to set up a Riemann sum an integrate. See if you can do this and send me your solution and/or questions.