hw3_sol_s11

# hw3_sol_s11 - Phys 2101 Homework 3 Solution Spring `11...

This preview shows pages 1–4. Sign up to view the full content.

Phys 2101 Homework 3 Solution Spring `11 These solutions use the parameter values from the problems printed in the book, not those that appear in your personal homework assignment. The logic used to get to the answer is the same, however. 1. 1. (a) From Table 2-1, we have v v a x 2 0 2 2 . Thus,       2 2 7 15 2 7 0 2 2.4 10 m/s 2 3.6 10 m/s 0.035 m 2.9 10 m/s. v v a x   (b) The initial kinetic energy is    2 2 27 7 13 0 11 1.67 10 kg 2.4 10 m/s 4.8 10 J. 22 i K mv  The final kinetic energy is    2 2 27 7 13 1.67 10 kg 2.9 10 m/s 6.9 10 J. f K mv The change in kinetic energy is K = 6.9 10 13 J 4.8 10 13 J = 2.1 10 13 J. 2. P. 7-13. We choose + x as the direction of motion (so a and F are negative-valued). (a) Newton’s second law readily yields 2 (85kg)( 2.0m/s ) F  so that 2 | | 1.7 10 N FF . (b) From Eq. 2-16 (with v = 0) we have     2 0 2 37m/s 0 2 3.4 10 m 2 2.0m/s v a x x     . Alternatively, this can be worked using the work-energy theorem. (c) Since F is opposite to the direction of motion (so the angle between F and d x   is 180°) then Eq. 7-7 gives the work done as 4 5.8 10 J W F x      .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
(d) In this case, Newton’s second law yields     2 85kg 4.0m/s F  so that 2 | | 3.4 10 N FF . (e) From Eq. 2-16, we now have     2 2 2 37m/s 1.7 10 m. 2 4.0m/s x    (f) The force F is again opposite to the direction of motion (so the angle is again 180°) so that Eq. 7-7 leads to 4 5.8 10 J. W F x      The fact that this agrees with the result of part (c) provides insight into the concept of work. 3. P. 7-14. The forces are all constant, so the total work done by them is given by W F x net , where F net is the magnitude of the net force and x is the magnitude of the displacement. We add the three vectors, finding the x and y components of the net force: net 1 2 3 net 2 3 sin50.0 cos35.0 3.00 N (4.00 N)sin35.0 (10.0 N)cos35.0 2.13N cos50.0 sin35.0 (4.00 N) cos50.0 (10.0 N)sin35.0 3.17 N. x y F F F F F F F            The magnitude of the net force is 2 2 2 2 net net net (2.13 N) (3.17 N) 3.82N. xy F F F The work done by the net force is net (3.82N)(4.00m) 15.3 J W F d where we have used the fact that d F net || (which follows from the fact that the canister started from rest and moved horizontally under the action of horizontal forces the resultant effect of which is expressed by F net ).
4. P. 7-17. We use F to denote the upward force exerted by the cable on the astronaut. The force of the cable is upward and the force of gravity is mg downward. Furthermore, the acceleration of the astronaut is a = g /10 upward. According to Newton’s second law, the force is given by 11 () 10 F mg ma F m g a , in the same direction as the displacement. On the other hand, the force of gravity has magnitude g F and is opposite in direction to the displacement.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

### Page1 / 12

hw3_sol_s11 - Phys 2101 Homework 3 Solution Spring `11...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online