Information
AI Chat

Physics Practice Problems

PHYS 103 practice with Prof. De about work, energy and simple harmonic motion

Course

General Physics I (PH 161)

63 Documents

Students shared 63 documents in this course

University

Montgomery College

Academic year: 2022/2023

Listed bookPhysics for Scientists and Engineers: a Strategic Approach

Uploaded by:

Anonymous Student

This document has been uploaded by a student, just like you, who decided to remain anonymous.

Montgomery College

Recommended for you

Comments

Please sign in or register to post comments.

Preview text

A 0. 150kg baseball is thrown straight up in the air w/ an initial velocity of 20 m/s . How high does the ball travel? ✓o = 20-0 V2 = V02 t 2. g - dy Uf : 0 Sy = ? by = t?---%ca?,----uoo- = a = 4. 19. 1- = ? m= 0. U& u can do it The Max speed of a child on a swing is 4. the Childs height above the ground is 0 at the lowest point of his motion. How high is he at his highest point ? ✓◦ = KE : PE Vf= 4. 9 DX = 0 42mi = Mgh a-_ 9. 1- = f- = F. DX. cos I 150 - cos ° = 9 40 → 150N Um

2° K°→F=io g- W= (207/10) Cosizo = -100 W= ( lo ) / SB) ( 0860 25 ⑤ 300N 1 W= 1300 ) (c ) COSO ¥ Isom 1%-0= F-- w= (lo ) (s ) cos /SO = - 20 -5m 1135° > 20N g (Zo) (5)cos 1350

70- 7/450 (Zo ) (5) Cosas = 70- 5m Spring = Fs = - 15 ✗= 0. 225 - 0 = -0. 125 (17-5)/-0-125 )

Lecture Slide Problems

What is the KE of a 1500kg car moving at 15 m/s? Yz. 1500 - 152

15 . 15 168,750J If the car changed its speed to 30m /S , how would the value of its KE?
I 500 - 30 - 30 675,000J

The human knee ligament can be a spring w/ a constant of 1h = 150 him. Whats the energy stored in the ligament if strained by 0 -75cm?fs = - KX450 ) (0-5) = 1 ?
A 0 ball is thrown straight up in the air w/ an initial velocity of 20. How high does the ball travel? M= 0 V2 = V02+ 2. g. Sy V0= 20. Vf= 0 Ay=%_gVI sy=? Ay

_ %%a = 20. " A =-9. t=

LoS0= 1 Exams review - in class 20590 = 0 COS180=- Energy in exam 3: work, power, energy, SHW PE= 0-o w = F.Sx

"

Yam = (330)(2) Cos o NE = 1/ mv 330. = 7005 PEs=Y2kxZ 350N Fs = - kx not horizontal force, w= F. COSO

I

asotewesoox

bro

ocoucometers

Whet = AKE T= 2 πE #x Fcoses 3320525: 31 k= lEx)

itis

31.7.50 = 1 31.7. 1 F='I2πT 15832 p= 35525 = 14. 14.7.50 = 0 3) OFN

####### eogI

w= F,T = AX = FCOSO lifting Up = 25 . negative + 0 = 0 (1) C work 0. 9.

25M
- T 2255 x = F= cosx = a = an (1)(29)
700N FOON ecoos x= Cosa"000cosa= = 4m

T 28005 2800 Fo aF = 2 = 10 Ls = COSOSX 0 8000N

I 98 ↓ esm 8008N W= g0xworkby 25m car F =c cannot be solved for billy Energy Problems
L00ng PE=mgh A 1000m (200) (9)(100) = 1,960, B Kt= 0, because vio it stops at the top, no speed A c)PE= (200) (a)(S00) rool or or

18/m/ as 135 What? Wret: AHE=KGo-KE. 0-112. - 2505

Work 16) 0.75.9.8.0.35=2
Energy 3)Kinetic 17 m = 1.9.8=8, at floor ) potential
law of conservation of energy 18)w= F.cos0 C) mechanical energy 350. = 700
Power 8)w = F.CO50 19) w= F.C888 w= (20) (050 = c105 3 5 - 50 = 1750 9)p= 2 500 = 100 x 3 = 300W 20x= lost

KE = 12mm 12.0.06=75J
PE = mgb 55.9.8=14653J (2)w= F.cos0 1x= W 2FcosO =

(3) (12.33= 17605 (4)(/2.2900 = 4,386,

10.9.8= 1965

1)w= F.C080 330-1080. 2: 700 2)w= F.C060 33 50 = 1389J 3)w= F.COs0 2 w = 0x = (0)(C =- 1. ( Ax FcOS o = 0. F = T k = 20V/m F = 20.0 = 5 1x = 0. x = 0 = 30 10 m = 0. SOU

Was this document helpful?