PHY 121 Exam2 Extra Review-thing students forget PDF

Preview

Summary

Download PHY 121 Exam2 Extra Review-thing students forget PDF

Description

1

2

3

Answer: 2

4

Answer: 4

5

Answer: 3

6

7

a. V = 9.80 m/s b. at pt P: v = 8.96 m/s Normal inward and gravity straight down.

c. Should have d. 100.9 m/s^2

8

9

a) Upward force for constant speed is 21640 N, so P = 64920 W (65kW) for this acceleration is 23440 N, so P = 46880 W (~47 kW)

b) Upward force

10

a) Gravity points straight down, Normal points perpendicular to track and is LARGER than Gravity, STATIC friction points inward down the track. b) Max speed is 26.8 m/s. NOTE THAT THIS IS MOST EASILY SOLVABLE IF you use a coordinate system where x-dir is horizontal and y-dir is vertical, NOT parallel and perpendicular to the track!!

11

See answers on the next slide.

12

13

a) Forces on 5 kg block: x-dir  left and , right; y-dir 5 down and  upward Forces on 10 kg block: x-dir , left, , left, 45.0 N pull to right y-dir: 10 down,  down and ; upward b) Tension: 9.8 N acceleration of 10.0 kg block (x-dir) 0.58 m/s^2

14

Solution on next slide

15

16

a)  =  − 4 =

 #

" !

− 5mg

b) two forces: Normal and mg point straight down d)  =

c)

5&

17

3. (a) Force applied by man is about 230 N (b) Work done by man is about -345 J (c) Work done by gravity is about 398 J (d) Work done by surface (friction) is about -53 J (e) 0 joules (no net force!) (f) 0 joules (no change in velocity)

18

a. initial vsystem = 5.08 m/s and thus the height reaches about 1.32 m b. initial vbullet = 980 m/s

19...