Work, vectors, and the inner product

Problem 4

Let's first compute how much work the force performs on the object as it moves from x = 0 meters to x = 3 meters. This is given by the integral of F dx. From x = 0 meters to 2 meters F is positive and the contribution to the integral is the area enclosed by the x-axis, the y-axis and the graph of the force which is 8 J.

From x = 2 meters to 3 meters the contribution to the integral is minus the area anclosed by the graph, the segment on the x-axis from x = 2 meters to x = 3 meters and the line parallel to the y axis given by x = 3 meters. This contribution is thus minus 2 J.

So, the total gain in kinetic energy equals 8 J - 2 J = 6 J. The kinetic energy at x = 0 meters was

1/2 * 1.5 kg (4.0 m/s)^2 = 12 J

So, it follows that the kinetic energy at x = 3 meters is 12 J + 6 J = 18 J which is answer a)

Problem 5

If A is a vector then by A^2 one means A dot A = |A|^2, the square of the length of the vector A.

In general, one can expand (A + B)^2 as follows:

(A + B)^2 = A^2 + B^2 + 2 A dot B

So, we can compute the magnitude of A + B if we know the magnitudes of A and B and the scalar product of A and B.
In the problem it is given that A^2 = 100 and B^2 = 121 and A dot B = -100, so we find that:

(A + B)^2 = 21, so the magnitude of A + B is sqrt(21) which is approximately 4.6, answer b).

Problem 6

You can solve this problem by writing C = p i-hat + q j-hat. Note that it is given that the z-component of C is zero, so you know that C is of this form. Then, from the given scalar products, you can solve for p and q and then compute the magnitude of C from that. However, this problem can also be solved without solving for C first. This works as follows. From A and B you construct a new orthonormal basis using the Gram-Schmidt process, see here:

http://en.wikipedia.org/wiki/Gram-Schmidt_process

Let's take our first basis vector, e1, to be in the direction of A. Then we have to normalize e1 such that it has a magnitude of 1, so we put:

e1 = A/|A| = A/sqrt(80)

Then, using B, we construct a vector orthogonal to A by subtracting from B the component of B in the direction of A. Let's call this vector u2:

u2 = B - (B dot e1) e1 = B - (B dot A)/A^2 A = B - 8/80 A = B - A/10

Here I used that the scalar product of A and B is 8 (A = 4 i + 8 j, 6 i - 2 j, so A dot B = 4*6 - 8*2 = 24 - 16 = 8).

The next step is to normalize u2. We put:

e2 = u2/|u2| = (B - A/10)/| B - A/10|

We can simplify this:

(B - A/10)^2 = B^2 + A^2/100 - A dot B/5 = 40 + 0.8 - 1.6 = 39.2

So, we have:

e2 = (B - A/10)/sqrt(39.2)

Now, we can expand C in terms of the new orthogonal unit vectors e1 and e2 using the general formula:

C = (C dot e1) e1 + (C dot e2) e2.

We have:

C dot e1 = C dot A/sqrt(80) = -16/sqrt(80)

And, using the linearity of the scalar product:

C dot e2 = [C dot B - 1/10 C dot A]/sqrt(39.2) = [18 + 16/10]/sqrt(39.2) = 19.6/sqrt(39.2)

The square of the magnitude of C is given by the sum of the squares of these inner products (just like when you express a vector in terms of the usual unit vectors i and j). To see this, just write

C^2 = [(C dot e1) e1 + (C dot e2) e2]^2, and expand using that e1^2 = e2^2 = 1 and e1 dot e2 = 0, yielding:

C^2 =(C dot e1)^2 + (C dot e2)^2 = 16^2/80 + 19.6^2/39.2 = 13. This means that:

|C| = sqrt(13) = approximately 3.6 = answer c)

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