7p42 we solve the problem using the work kinetic

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Unformatted text preview: rectangular “areas” in the graph (for 0 ≤ x ≤ 4) gives 42 J for the work done. (b) Counting the “areas” under the axis as negative contributions, we find (for 0 ≤ x ≤ 7) the work to be 30 J at x = 7.0 m. (c) And at x = 9.0 m, the work is 12 J. (d) Equation 7- 10 (along with Eq. 7- 1) leads to speed v = 6.5 m/s at x = 4.0 m. Returning to the original graph (where a was plotted) we note that (since it started from rest) it has received acceleration(s) (up to this point) only in the +x direction and consequently must have a velocity vector pointing in the +x direction at x = 4.0 m. (e) Now, using the result of part (b) and Eq. 7- 10 (along with Eq. 7- 1) we find the speed is 5.5 m/s at x = 7.0 m. Although it has experienced some deceleration during the 0 ≤ x ≤ 7 interval, its velocity vector still points in the +x direction. (f) Finally, using the result of part (c) and Eq. 7- 10 (along with Eq. 7- 1) we find its speed v = 3.5 m/s at x = 9.0 m. It certainly has experienced...
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This document was uploaded on 03/24/2014.

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