11 please easy way to solve and use this formula sheet please v2Circular motion: a =RFnet = ma;Weight: Fg = mg, ;g = 9.8 m/s?;Payr =At1(Kinetic energy: K = –m v²;2Potential energy: Ug=mgyE = ErE = U+Kv2Rotational motion:1 rev = 2n rad;v = o r;= 0?r;at = a r;a, =1K=-lo2@ = 00 + at;0 = 0o t+22θα -ω -ω=rxFT= rFsino;1Ipoint massIgF mRΣτ-Ια;= mr?Idisk =mR+%3D22112Iball = =mR?2Ishell =3 MR2Ihoop = mR?Irod(center) =mL?Irod(end) =12mL?work: W=r0; K=-lo ?dWP =dtWPavr =AtI= Icom + MD?W =22Rolling:K= lo221m VcomVcom = RoT= f,RF,max = HFnIncline: F=mgsin0 F=mgcos0Angular momentum:Lpoint mass =m rxvL=mrvsin 0;L=m (r,Vy - ryVz)kL= IoLI= LI 01 =I 202m,x1 +m2X2miy1+m2y2X comY comm1 +m2m1 +m21 B) 9 m/sC) 16 m/sD) 25 m/sE) >25 m/s11. A 2-kg block starts from rest at the top of a 10-m long ramp inclined at 45 deg to the horizontal. The block is releasedand begins to slide down the ramp. At the bottom of the ramp the block has a speed of 9 m/s. What is the coefficient offriction between the block and the ramp?A) 0.12B) 0.22C) 0.32D) 0.42E) 0.52

11. A 2-kg block starts from rest at the top of a 10-m long ramp inclined at 45 deg to the horizontal. The block is released and begins to slide down the ramp. At the bottom of the ramp the block has a speed of 9 m/s. What is the coefficient of friction between the block and the ramp? A) 0.12 B) 0.22 C) 0.32 D) 0.42 E) 0.52 4

11. A 2-kg block starts from rest at the top of a 10-m long ramp inclined at 45 deg to the horizontal. The block is released and begins to slide down the ramp. At the bottom of the ramp the block has a speed of 9 m/s. What is the coefficient of friction between the block and the ramp? A) 0.12 B) 0.22 C) 0.32 D) 0.42 E) 0.52 4

Principles of Physics: A Calculus-Based Text

5th Edition

ISBN:9781133104261

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter7: Conservation Of Energy

Section: Chapter Questions

Problem 64P

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Concept explainers

Centripetal Force

A particle, body, object or certain mass while travelling in a rotational path or semicircular path, experiences an inertia force. If that inertia force pulls the towards the inward of the imaginary circle in the rotation, it is termed as centripetal force.

Question

11 please easy way to solve and use this formula sheet please

v2
Circular motion: a =
R
Fnet = ma;
Weight: Fg = mg, ;
g = 9.8 m/s?;
Payr =
At
1
(Kinetic energy: K = –m v²;
2
Potential energy: Ug=mgy
E = Er
E = U+K
v2
Rotational motion:
1 rev = 2n rad;
v = o r;
= 0?r;
at = a r;
a, =
1
K=-lo
2
@ = 00 + at;
0 = 0o t+
2
2θα -ω -ω
=rxF
T= rFsino;
1
Ipoint mass
IgF mR
Στ-Ια;
= mr?
Idisk =
mR
+
%3D
2
2
1
1
2
Iball = =mR?
2
Ishell =
3 MR2
Ihoop = mR?
Irod(center) =
mL?
Irod(end) =
12
mL?
work: W=r0; K=-lo ?
dW
P =
dt
W
Pavr =
At
I= Icom + MD?
W =
2
2
Rolling:
K= lo
2
2
1
m Vcom
Vcom = Ro
T= f,R
F,max = HFn
Incline: F=mgsin0 F=mgcos0
Angular momentum:
Lpoint mass =m rxv
L=mrvsin 0;
L=m (r,Vy - ryVz)k
L= Io
LI= L
I 01 =I 202
m,x1 +m2X2
miy1+m2y2
X com
Y com
m1 +m2
m1 +m2
1

B) 9 m/s
C) 16 m/s
D) 25 m/s
E) >25 m/s
11. A 2-kg block starts from rest at the top of a 10-m long ramp inclined at 45 deg to the horizontal. The block is released
and begins to slide down the ramp. At the bottom of the ramp the block has a speed of 9 m/s. What is the coefficient of
friction between the block and the ramp?
A) 0.12
B) 0.22
C) 0.32
D) 0.42
E) 0.52

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