Egg Drop Tutorial
Drop vs. Stop
Mass of egg = 0.0600 kg (60 grams)
Average time of stopwatches = _ s (get the times from your teacher: t 1 = _s t2 = _s t3 = _s)
Before you start this tutorial, remind yourselves of how v-t graphs work and why they are impo
Solutions: Phy 11, HW 1, Spring
1.10.
IDENTIFY: Convert units.
SET UP: Use the unit conversions given in the problem. Also, 100 cm 1 m and 1000 g 1 kg.
ft
mi 1 h 5280 ft
EXECUTE: (a) 60
88
h 3600 s 1 mi
s
m
ft 3048 cm 1 m
(b) 32 2
98 2
s
s 1 ft
Solutions: Phy 11, HW 2, Spring
2.19.
IDENTIFY: Use the constant acceleration equations to find v0x and ax
(a) SET UP: The situation is sketched in Figure 2.19.
x x0 700 m
t 700 s
vx 15.0 m/s
v0 x ?
Figure 2.19
2( x x0 )
2(700 m)
v vx
vx
150 m/s 50
Solutions: Phy 11, HW 5, Spring
4.2.
IDENTIFY: We know the magnitudes and directions of three vectors and want to use them to find their components, and
then to use the components to find the magnitude and direction of the resultant vector.
SET UP: Let F1
Solutions: Phy 11, HW 6, Spring
5.8.IDENTIFY: Apply Newtons first law to the car.
SET UP: Use x and y coordinates that are parallel and perpendicular to the ramp.
EXECUTE: (a) The free-body diagram for the car is given in Figure 5.8. The vertical weight w
Solutions: Phy 11, HW 7, Spring
5.74.
IDENTIFY: Apply F ma to the brush. Constant speed means a 0. Target variables are two of the forces on the
brush.
SET UP: Note that the normal force exerted by the wall is horizontal, since it is perpendicular to the
Solutions: Phy 11, HW 8, Spring
5.48.IDENTIFY: Apply F = ma to the button. The button moves in a circle, so it has acceleration arad .
SET UP: The situation is equivalent to that of Example 5.21.
2
2 R
v2
EXECUTE: (a) s
so s 42 R . A platform speed of 40
Physics 11
Introductory Physics I Spring 2016
Department of Physics, Lehigh University
Michael Stavola
(610) 758-3946
E-mail: mjsa
Room 211, Sherman Fairchild Laboratory.
Office Hours: Tues., 1:30 pm 5:30 pm
Recitation Instructor and Section Number: _
Con
Sample Exam, Spring 2015
Physics 11, Hour exam 1, February 17, 2014
NAME: _
Recitation Instructor and Section Number: _
Exam is closed book. You may use the equation sheet on the back of this exam and a
calculator. No other notes are permitted.
All work o
1) Charlie heaves a 7.25 kg shot straight upward, giving it a constant upward acceleration from rest of
35.0 m/s2 for 0.640 m. Charlie releases it 2.20 m above the ground. (Hint: You might like to draw a
figure to keep track of the shot’s elevation at var
1.
A50.0kgskierstartingfromresttravels200mdownahillthathasa20.0slopeandauniform
surface.Whentheskierreachesthebottomofthehill,herspeedis30.0m/s.
a. Howmuchworkisdonebyfrictionastheskiercomesdownthehill?
b.
Whatisthemagnitudeofthefrictionforceiftheskiertra
FORCE is a push or pull due to interaction between two obje ~ r.\
W. (magnitude and direction, 3 VE
Checklist of fiVe common types 0 orces we will consider in this course:
[:1 The only non-contact force in Physics 11 is WEIGHT. It is a pull on an
object
Physics 11, Lecture 7 Outline, 9/15/2014
Dynamics on Curved Paths
What path will ball follow when it reaches
gap in ring?
Angular motion: Sect 9.1 Us 5. rdinates.
Measure 9 count(erclockwi)som pos. x-axis. (Fig. 9.11)
- Quivr c Mom :
Consrder stepped pu
SVW'Q7/ ReSOVVCM/ Midjibé/ ad/m
Ph sics 11 Lecture 10 Outline 9/24/2014 Potential Ener
)Recall from last time, workenergy theorem. (Eqn 6.6) W ,
'7 - -
'1:er : cliqng I- (mt/712ean " in
I Section 7.1: Gravity Rock at top of hill has some potential or sto
Physics 11, Lecture 12 October 1, 2014 Energy Wrapup
Caution: Gravitational S stem \
In several places, we, have received cautionary notes that for gravity, both \0"
objects have to be treated as part of the system. Sect 4.5, p 121: When 9
$0 you drop t
Lecture 15 October 15, 2014 Rotation *
REFRESHER (from Centripetal Acceleratio discussion from Ch - er 3
Angular motion: (Sect 9.1) Use trig cooyin/arlces/Measur .
/
( l .
from pos. x-axis. (Fig. 9.11)
Recall stepped pulley, s E arclength (Eqn 9.1). Use
K Cow-v1 AV 3 A
CGHISW$-' P °°"5°"/ 6/042 MS 0'1
Lfm.7lrl Reyes-AM 9cm OLA-w
Lecture 14 October 13, 2014 Center of Mass Last time: p = 71117 , impulse, collision.
Section 8.5 Center of M
a
For a syslgn 0}pnicle For m1 (X1 {1), m2 at (x2,y2), etc.
/_ mlx1