1. No category

GENERAL PHYSICS PH 221-3A (Dr. S. Mirov)
Test 2 (10/10/07)
Sample Test 2
STUDENT NAME: _Key____________________ STUDENT id #: ___________________________
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ALL QUESTIONS ARE WORTH 20 POINTS. WORK OUT FIVE PROBLEMS.
NOTE: Clearly write out solutions and answers (circle the answers) by section for each part (a., b., c., etc.)
Important Formulas:
1. Motion along a straight line with a constant acceleration
vaver. speed = [dist. taken]/[time trav.]=S/t;
vaver.vel. = ∆x/∆t;
vins =dx/∆t;
aaver.= ∆vaver. vel./∆t;
a = dv/∆t;
v = vo + at; x= 1/2(vo+v)t; x = vot + 1/2 at2; v2 = vo2 + 2ax (if xo=0 at to=0)
2.
Free fall motion (with positive direction ↑)
g = 9.80 m/s2;
y = vaver. t
vaver.= (v+vo)/2;
v = vo - gt; y = vo t - 1/2 g t2; v2 = vo2 – 2gy (if yo=0 at to=0)
3.
Motion in a plane
4.
Projectile motion (with positive direction ↑)
5.
Uniform circular Motion
vx = vo cosθ;
vy = vo sinθ;
x = vox t+ 1/2 ax t2; y = voy t + 1/2 ay t2; vx = vox + at; vy = voy + at;
vx = vox = vo cosθ;
x = vox t;
xmax = (2 vo2 sinθ cosθ)/g = (vo2 sin2θ)/g for yin = yfin;
vy = voy - gt = vo sinθ - gt;
y = voy t - 1/2 gt2;
a=v2/r,
T=2πr/v
6. Relative motion



v=
v PB + v BA
PA


a PA = a PB
7.
Component method of vector addition
A=
Ax2 + Ay2 ; θ = tan-1 Ay /Ax;


The scalar product A a ⋅ b =ab cos φ


a ⋅ b= ( a x iˆ + a y ˆj + a z kˆ ) ⋅ ( bx iˆ + by ˆj + bz kˆ )
 
a ⋅ b =a x bx + a y by + a z bz


The vector product a × b= ( a x iˆ + a y ˆ
j + a z kˆ ) × ( bx iˆ + by ˆj + bz kˆ )
A = A1 + A2 ; Ax= Ax1 + Ax2 and Ay = Ay1 + Ay2;
kˆ
iˆ




a × b =−b × a =a x
ˆj
ay
a z =iˆ
bx
by
bz
ay
az
by
bz
− ˆj
ax
az
bx
bz
+ kˆ
ax
ay
bx
by
=
= ( a y bz − by a z )iˆ + ( a z bx − bz a x ) ˆj + ( a x by − bx a y ) kˆ
1.
Second Newton’s Law
ma=Fnet ;
2.
Kinetic friction fk =µkN;
3.
Static friction fs =µsN;
4.
Universal Law of Gravitation: F=GMm/r2; G=6.67x10-11 Nm2/kg2;
5.
Drag coefficient
6.
Terminal speed
7.
Centripetal force: Fc=mv2/r
8.
Speed of the satellite in a circular orbit: v2=GME/r
9.
The work done by a constant force acting on an object:
10. Kinetic energy:
D =
vt =
K =
1
C ρ Av 2
2
2mg
Cρ A
W=
Fd cos φ=
1
mv 2
2
11. Total mechanical energy: E=K+U
12. The work-energy theorem: W=Kf-Ko; Wnc=∆K+∆U=Ef-Eo
13. The principle of conservation of mechanical energy: when Wnc=0, Ef=Eo
14. Work done by the gravitational force:
Wg = mgd cos φ
 
F ⋅d
1.
Work done in Lifting and Lowering the object:
∆K =
K f − Ki =
Wa + Wg ; if K f =
K i ; Wa =
−Wg
Fx = − kx (Hook's law)
2.
Spring Force:
3.
Work done by a spring force:
1
1
1
Ws =
kxi2 − kxo2 ; if xi =
x; Ws =
− kx 2
0 and x f =
2
2
2
xf
4.
Work done by a variable force:
W =
∫ F ( x )dx
xi
Pavg=
W
; P=
∆t
dW
; P=
dt
5.
Power:
6.
Potential energy:
7.
Gravitational Potential Energy:
∆U =
Fv cos φ=
∆U =
−W ; ∆U =
−∫
mg ( y f − yi ) =
xf
xi
 
F ⋅v
F ( x )dx
mg ∆y ; if yi = 0 and U i = 0; U ( y ) =
mgy
1
kx 2
2
8.
Elastic potential Energy:
U ( x) =
9.
Potential energy curves:
dU ( x )
F ( x) =
; K ( x) =
Emec − U ( x )
−
dx
10. Work done on a system by an external force:
Friction is not involved W = ∆Emec = ∆K + ∆U
When kinetic friction force acts within the system W = ∆Emec + ∆Eth
∆Eth =
fk d
11. Conservation of energy:
12. Power:
=
Pavg
∆E
=
; P
∆t
W = ∆E = ∆Emec + ∆Eth + ∆Eint
for isolated system (W=0) ∆Emec + ∆Eth + ∆Eint =0
dE
dt
1.
4T
4T
61.2 kg
5
2
-2
=4x103 m
6.
7.
(a)
(b)
(c)
(d)
A 700 g block is released at height ho above a vertical spring of constant k=400
N/m and negligible mass. The block sticks to the spring and momentarily stops
after compressing the spring 19.0 cm. How much work is done
by the block on the spring?
by the spring on the block?
What is the value of ho?
If the block were released from height 2.00 ho above the spring, what would be the
maximum compression of the spring?
We place the reference position for evaluating gravitational potential energy at the
relaxed position of the spring. We use x for the spring's compression, measured positively
downwards (so x > 0 means it is compressed).
(a) With x = 0.190 m,
Ws = −
1
kx 2 = −7.22 J ≈ −7.2 J
2
for the work done by the spring force. Using Newton's third law, we see that the work
done on the spring is 7.2 J.
(b) As noted above, Ws = –7.2 J.
(c) Energy conservation leads to
Ki + U i = K f + U f
mgh0 = − mgx +
1
kx 2
2
which (with m = 0.70 kg) yields h0 = 0.86 m.
(d) With a new value for the height h0′ = 2h0 = 1.72 m , we solve for a new value of x
using the quadratic formula (taking its positive root so that x > 0).
mg +
1
mgh0′ = − mgx + kx 2 ⇒ x =
2
which yields x = 0.26 m.
bmg g
k
2
+ 2mgkh0′
8. A particle moving along the x axis is acted upon by a single force F = F0e-kx,
where F0 and k are constants. The particle is released from rest at x = 0. What is
the maximum kinetic energy it will attain?
According to the work energy theorem W = K f − Ki = K f since we started from rest (Ki = 0)
f
− kx
 e − kx f 1 


e
− kx
On the other hand W = ∫ F ( x )dx → K = ∫ F ( x )dx = Fo ∫ e dx = Fo  −
 =Fo  − k + k 
0
0
0
k

0


− kx
Max K f occurs at a point when=
K ' ( x ) 0,=
K ' ( x ) F=
) 0 only for x=∞
o (e
x
xf
Fo
→ K max =
Joules
k
xf
xf
9. The potential energy of a 0.20-kg particle moving along the x axis is given by
U(x) = (8.0J/m2)x2 + (2.0J/m4)x4 . When the particle is at x = 1.0m it is traveling in the positive x
direction with a speed of 5.0m/s. At what x coordinate it will stop momentarily to turn around.
( a ) Calculate total mechanical energy of the particle at coordinate x=1.0 m
mv 2 0.2 × 5.02
U (1.0m ) = 8.0 J + 2.0 J = 10.0 J ; K (1.0m ) =
=
= 2.5J
2
2
→ Emec =
12.5J
(b) The particle will stop momentarily to turn around when K ( x ) =Emec − U ( x ) =0
12.5
→ 8.0 x 2 + 2.0 x 4 =
−8.0 ± 8.02 + 4(2.0)(12.5)
x =
4.0
x2 =
( −2.0 ± 3.2)m 2
2
x 2 = 1.2m 2
x = 1.1m