04_InstSolManual_PDF_Part3

# 04_InstSolManual_PDF_Part3 - Solve v x 5 v x 1 a x t 5 1...

This preview shows page 1. Sign up to view the full content.

and its components are shown in Figure 4.5. Figure 4.5 Reflect: Adding the forces as vectors gives a very different result from adding their magnitudes. 4.6. Set Up: Let be the direction of the force and acceleration. Solve: gives 4.7. Set Up: Let be the direction of the force. Use a constant acceleration equation to calculate the displacement of the probe. Solve: (a) (b) 4.8. Set Up: Take to be the direction in which the skater is moving initially. (comes to rest). Solve: so The only horizontal force on the skater is the friction force, so The force is 46.7 N, directed opposite to the motion of the skater. 4.9. Set Up: Take to be in the direction in which the cheetah moves. Solve: (a) so (b) The force is exerted on the cheetah by the ground. Reflect: The net force on the cheetah is in the same direction as the acceleration of the cheetah. 4.10. Set Up: Let be the direction of the force. Use to calculate the acceleration of the puck. Then use constant acceleration equations to find the speed and displacement of the puck for
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Solve: v x 5 v x 1 a x t 5 1 1.56 m / s 2 21 2.00 s 2 5 3.12 m / s x 2 x 5 v x t 1 1 2 a x t 2 5 1 2 1 1.56 m / s 2 21 2.00 s 2 2 5 3.12 m v x 5 0. a x 5 g F x m 5 0.250 N 0.160 kg 5 1.56 m / s 2 . t 5 2.00 s. g F x 5 ma x g F x 5 0.250 N. 1 x F x 5 ma x 5 1 68 kg 21 10.05 m / s 2 2 5 680 N a x 5 v x 2 v x t 5 20.1 m / s 2 2.0 s 5 10.05 m / s 2 v x 5 v x 1 a x t v x 5 0. 1 x f x 5 ma x 5 1 68.5 kg 21 2 0.682 m / s 2 2 5 2 46.7 N. a x 5 v x 2 v x t 5 2 2.40 m / s 3.52 s 5 2 0.682 m / s 2 v x 5 v x 1 a x t v x 5 1 x x 2 x 5 v x t 1 1 2 a x t 2 5 1 2 1 4.31 m / s 2 21 10.0 s 2 2 5 215 m t 5 10.0 s. a x 5 4.31 m / s 2 , v x 5 0, a x 5 g F x m 5 140 N 32.5 kg 5 4.31 m / s 2 x 2 x g F x 5 140 N. 1 x m 5 g F x a x 5 48.0 N 3.00 m / s 2 5 16.0 kg. g F x 5 ma x g F x 5 48.0 N. 1 x u R R y y R x x O R S Newton’s Laws of Motion 4-3...
View Full Document

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern