Transcript Ch-8

Potential energy is energy which results from position or
configuration.
The negative sign provides the convention that work done against a force field
(gravitational field) increases potential energy, while work done by the force field
(gravitational field) decreases potential energy.
1.
A conservative force is a force with the property that the
work done in moving a particle between two points is
independent of the taken path.
2.
Equivalently, if a particle travels in a closed loop, the net
work done (the sum of the force acting along the path multiplied
by the distance travelled) by a conservative force is zero.
Examples: Conservative forces, the gravitational force, spring force, electric force
Non conservative forces friction and air drag
No, (consider the round trip
on the small loop)
U= -Fxdx
Calculate area under the curve
1) U=-(Fi X1)/2
2) U=-(Fi X1)
3) U=-[-(Fi X1)/2]= FiX1/2
Ans: 3, 1, 2
3, 1, 2
(a) All tie, (b) all tie
(t = 0 s), (m = 1 kg), and (v = 24 m/s j)
Solving this equation for a, substituting the result into Eq. 8-27
All tie
General: The definition of law of conservation of
energy is stating that energy cannot be created or
destroyed; but, it can be changed or transferred
from one form to another.
Physics: The law stating that the total amount
of energy in any isolated system remains constant,
and cannot be created or destroyed, although it may
change forms.
h
Ui
Uf and the kinetic energy is zero both
at the beginning and end. Since energy is conserved
kx2+-2mgx-2mgh=0
https://www.youtube.com/watch?v=cszUKo8J7aI
(1)
(2)
(a)
(b)
(c)
k
From 1, fk = 7.42 N
=
Thus, the change in height between the
car's highest and lowest points is
Problem : What is the total energy of the mass-spring system shown below? The mass is
shown at its maximum displacement on the spring, 5 meters from the equilibrium point.
Here is system of two conservative forces, mass and gravity.
Ug = mgh = - 5mg = - 245 Joules. Also, Us = 1/2kx2 =
1/2(10)(5)2 = 125 Joules. Thus the total potential energy, and
hence the total energy is the sum of these two quantities: E =
Ug+Us = - 120 Joules.
Problem 3: A ball of mass 2.00-kg is dropped from a height of 1.5 m (from
the ground) onto a massless spring (the spring has an equilibrium length of
0.50 m). The ball compresses the spring by an amount of 0.20 m by the
time it comes to a stop. Calculate the spring constant of the spring.
Initial Energy=Final energy
mg(1.0 m) = mg(-0.20 m) + ½ k(-0.20 m)2
(2.00 kg)(9.80 m/s2)(1.0 m) + (2.00 kg)(9.80 m/s2)(0.20 m)= 0.02 k
9.6 J + 3.92 J = 0.02 k
k = (23.5)/(0.02) = 1,180 N/m
Problem 2: A 200-kg crate is pulled along a level surface by an engine. The coefficient of friction between the
crate and the surface is mk = 0.40.
(a) How much power must the engine deliver to move the crate at 5.0 m/s?
Solution:
The engine must work against the frictional force, therefore it has to exert a force along the direction
of displacement equal to the frictional force. Therefore:
fk= mkmg = (0.40)(200 kg)(9.80 m/s2) = 784 N
P = F.v = (784 N)(5.0 m/s) = 3,920 Watts
(b) How much work is done by the engine in 3 min. (3 min = 180 s)?
Solution:
P = W/t therefore, W = Pxt
W = (3,920 J-s)x(180 s) = 706 kJ
PROBLEM: A 220 kg crate is lifted a distance of 21.0 m vertically with an acceleration of 1.47 m/s2 by a single cable.
(a) Find the tension on the cable
Solution:
The system is NOT in equilibrium since there is a net acceleration. Therefore:
T - mg = ma, T = m (g + a) = (220 kg)( 9.80 + 1.47)
T = 2,480 N
(b) Find the net work done on the crate
Solution:
The net work done is done by the net force: Fnet = ma = (220 kg)(1.47 m/s2) = 323 N
Wnet = (323 N)(21.0 m) = 6,800 J
(c) the work done by gravity on the crate
Solution:
Wg = (Fg cos 180)(21.0 m) = -45,300 J
(d) Find the work done by the cable on the crate
Solution:
WT = T d = (2480 N)(21.0 N) = 52,100 J
(e) find the final speed of the crate assuming it started from rest.
Solution:
The werk-energy theorem states that the change in the kinetic energy of the system is equal to the net wok
done. The system starts from rest, therefore the final kinetic energy = 0.
K = 1/2(220 kg)vf2
1/2(220 kg)vf2= 6800 J
vf2= 2(6800 J)/(220 kg) = 61.8
vf= 7,86 m/s