15. A 3-kg box is moving at 6.0 m/s atop a hill that is 5.0 m above a flat frictionless surface. It descends the hill (also frictionless) and at the bottom of the hill contacts an ideal spring, which compresses 1.53 m before stopping. What is the spring constant k?

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Author:Raymond A. Serway, Chris Vuille
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Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Do not copy from chegg use easy way to explain use provided formula please only 15
zt;
c d
a
A) a
B) b
C) с
D) d
E) in the region from 0 to a, and from c to infinity
Cen
15. A 3-kg box is moving at 6.0 m/s atop a hill that is 5.0 m above a flat frictionless surface. It descends the hill (also
frictionless) and at the bottom of the hill contacts an ideal spring, which compresses 1.53 m before stopping. What is the
spring constant k ?
A) 172 N/m
B) 225 N/m
C) 344 N/m
D) 450 N/m
E) 980 N/m
5
Lpoint mass
m rxv
I 01 =I 202
X com
Transcribed Image Text:zt; c d a A) a B) b C) с D) d E) in the region from 0 to a, and from c to infinity Cen 15. A 3-kg box is moving at 6.0 m/s atop a hill that is 5.0 m above a flat frictionless surface. It descends the hill (also frictionless) and at the bottom of the hill contacts an ideal spring, which compresses 1.53 m before stopping. What is the spring constant k ? A) 172 N/m B) 225 N/m C) 344 N/m D) 450 N/m E) 980 N/m 5 Lpoint mass m rxv I 01 =I 202 X com
W
Pavr =
At
v?
Weight: Fg=mg, ;
g = 9.8 m/s²;
2022
Fnet = ma;
Circular motion: a =
R
Vhat
1
m v²
Potential energy: Ug=mgy
E = E,
E = U+K
(Kinetic energy: K =
v2
1 rev = 2n rad;
- = o°r;
v = 0 r;
a = a r;
ar =
Rotational motion:
may b<
T=rFsino;
20a = 0? - @o?
K=
2
T=rxF;
0 = 0o + at;
e = 0o t+
2
1
mR 2
2
+Rin
Iaisk =mR2
pe =
Στ-Ια;
Ipointmass = mr?
2
Iball = =mR?
mR?
1
-mL?
Irod(end) =
Ishell =
Ihoop = mR?
Irod(center) =
12
W
dW
K- w- , P-
Payr =
At
2
2
W =
2
I= Icom + MD²
work: W=t0;
dt
K=
2+
m vcom 2
T= fR
F,max = H,Fn
Rolling:
Vcom = Rø
3-k
Jes:
suc
Incline: F=mgsin0 F=mgcos0
72 N
5N
N/
N/m
/m
L= mrvsin 0;
L=m (r,vy - ryVz)k
L= Io
Angular momentum:
Lpoint mass =m rxv
miy1+m2y2
m,X1 +m2x2
m1 +m2
LJ = L
I o =I 202
X com
Y com =
m1 +m2
1
Transcribed Image Text:W Pavr = At v? Weight: Fg=mg, ; g = 9.8 m/s²; 2022 Fnet = ma; Circular motion: a = R Vhat 1 m v² Potential energy: Ug=mgy E = E, E = U+K (Kinetic energy: K = v2 1 rev = 2n rad; - = o°r; v = 0 r; a = a r; ar = Rotational motion: may b< T=rFsino; 20a = 0? - @o? K= 2 T=rxF; 0 = 0o + at; e = 0o t+ 2 1 mR 2 2 +Rin Iaisk =mR2 pe = Στ-Ια; Ipointmass = mr? 2 Iball = =mR? mR? 1 -mL? Irod(end) = Ishell = Ihoop = mR? Irod(center) = 12 W dW K- w- , P- Payr = At 2 2 W = 2 I= Icom + MD² work: W=t0; dt K= 2+ m vcom 2 T= fR F,max = H,Fn Rolling: Vcom = Rø 3-k Jes: suc Incline: F=mgsin0 F=mgcos0 72 N 5N N/ N/m /m L= mrvsin 0; L=m (r,vy - ryVz)k L= Io Angular momentum: Lpoint mass =m rxv miy1+m2y2 m,X1 +m2x2 m1 +m2 LJ = L I o =I 202 X com Y com = m1 +m2 1
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