course phy201
et 3 Problem number 9--------------------------------------------------------------------------------
Problem
Solution
Generalized Solution
Problem
An object is moving at 17 meters per second.
If its mass is 29 kilograms, what is its kinetic energy?
If its speed doubled, what would be its kinetic energy?
How many times the original kinetic energy would it have after its speed doubled?
V0=17m/s
m= 29kg
Fnet= m * a
not sure?????
Solution
When v is 17 m/s and m is 29 Kg, we have
KE= .5( 29 Kg)( 17 m/s)^2 = .5( 29 Kg)( 17 m/s)^2 = 4190.5 Joules.
The same procedure, using 34 m/s, yields kinetic energy 16762 Joules.
The factor by which the kinetic energy increases is the ratio 16762/ 4190.5 = 4 of these results.
Generalized Solution
At velocity v, the kinetic energy of a mass m is KE1 = .5 m v^2.
At velocity v ' = 2v, the kinetic energy is
KE2 = .5 m (v ' )^2 = .5 m (2v) ^ 2 = .5 m (4 v^2).
The ratio of kinetic energies is therefore
KE2 / KE1 = .5 m (4 v^2) / (.5 m v^2) = 4.
More generally if velocity increases by factor c:
From v to v' = cv, the kinetic energy will increase from KE1 = .5 m v^2 to KE2 = .5 m (cv)^2 = .5 m (c^2 v^2), and the ratio of kinetic
energies will be KE2 / KE1 = c^2.
Kinetic energy thus increases by a factor equal to the square of the velocity increase.
As seen above doubling velocity doubles kinetic energy.
If velocity is tripled kinetic energy will increase by factor 3^2 = 9.
If velocity increases to 10 times its original value kinetic energy will increase by factor 100.
This, incidentally, is the reason why it is so much worse to run your car into a solid object at 60 mph than at 30 mph: the KE is four
times as great at 60 mph than at 30 mph.
It also takes 4 times as far to stop a car from 60 mph than from 30 mph.
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
OK, So we use the formula KE1= .5 m v^2
KE1= .5(29kg) * (17M/s)^2
Ke1= 14.5 kg * 289m^2 /s^2
Ke1= 4190.5 joules
Speed doubled 17m/s *2 = 34m/s
KE2= .5(29kg) * (34m/s)^2
KE2= 14.5kg * 1156m^2/s^2
KE2=16762joules
16762/4190.5= 4 times the original Ke
@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
Set 3 Problem number 10
--------------------------------------------------------------------------------
Problem
Solution
Generalized Solution
Explanation in terms of Figure(s), Extension
Figure(s)
Problem
An object, initially at rest, is acted upon by a net force of 11.5 Newtons.
The object has mass 2.3 kilograms.
The force acts for 4.7 seconds.
What velocity will the object attain and how far will it travel during this time?
What kinetic energy will it attain?
Fnet=11.5newtons
m=2.3kg
'dt=4.7s
V0=0
Fnet= m * a
11.5 kg m/s^2 = 2.3kg * a
a= 5 m/s^2
Vf= V0 + a * 'dt
Vf= 0 + 5m/s^2 * 4.7s
Vf= 23.5m/s
'ds = vave * 'dt
'ds= (23.5m/s + 0m/s)/2 * 4.7s
'ds= 55.2 meters
KE1= .5m * V^2
KE1= .5(2.3kg) * (23.5m/s)^2
KE1=1.15kg * 552.25m^2/s^2
KE1=635.09joules
Solution
The acceleration of an object of mass 2.3 Kg under the influence of a 11.5 Newton force will be a=F/m=( 11.5 Newtons)/( 2.3 Kg)= 5 meters
per second per second.
In 4.7 seconds the velocity change will be 4.7( 5 m/s) = 23.5 meters per second; since initial velocity is zero this will be the velocity
at the end of the 4.7 seconds.
The average velocity will be the average of this velocity and zero, or ( 23.5 + 0)/2 meters per second = 11.75 meters per second.
At this average velocity, in 4.7 seconds the object will move 55.225 meters.
The kinetic energy will be KE = .5( 2.3 Kg)( 23.5 m/s)^2 = 635.0875 Joules.
The work done will be the product of the 11.5 Newton force and the 55.225 meter displacement, or 635.0875 Joules.
Generalized Solution
If an object of mass m, initially at rest, is acted upon by a net force F for time interval `dt, it will experience acceleration a = F / m
for time interval `dt.
This will result in a velocity change `dv = a `dt = F / m * `dt.
Since the object started from rest its final velocity will be
vf = 0 + `dv = F / m * `dt
and its average velocity will be
vAve = (v0 + vf ) / 2 = (0 + F / m * `dt) / 2 = (1/2) (F / m) `dt.
The distance traveled by the object will be
`ds = vAve `dt = (1/2) (F / m) `dt * `dt = (1/2) (F / m) * `dt^2.
The kinetic energy attained by the object will be
KE = .5 m vf^2 = .5 m (F / m * `dt) ^ 2 = (1/2) F^2 / m * `dt^2.
Explanation in terms of Figure(s), Extension
The figure below shows the complete relationship between the work done by the net force and the kinetic energy gained by the object.
For the purposes of this problem, note that vf is obtained by the same series of relationships as before and ignore everything to the
right of v0.
Note that if we simply combine our knowledge of vf with the originally given mass m, we obtain KE = .5 m vf^2.
As determined earlier, if v0 = 0 then vf = F / m * `dt, so after a little substitution and simplification we will find that KE = (1/2) F^2
/ m * `dt^2.
Figure(s)
@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
Set 3 Problem number 11
--------------------------------------------------------------------------------
Problem
Solution
Generalized Solution
Explanation in terms of Figure(s), Extension
Figure(s)
Problem
An object of mass 7.3 kilograms is acted upon by a net force of 32.85 Newtons.
The object is initially at rest.
After the first 7.3 seconds, what will the its velocity, and how far will it have traveled?
How much work is done?
m=7.3kg
Fnet=32.85newtons
V0=0
'dt=7.3s
Fnet= M * a
32.85 newtons = 7.3kg * a
a= 4.5m/s^2
Vf=v0 + a * 'dt
Vf= 0 + 4.5m/s^2 * 7.3s
vf= 32.85m/s
'ds= vave * 'dt
'ds= (32.85m/s + 0m/s)/2 * 7.3s
'ds= 119.90meters
KE1= .5m V^2
KE1= .5(7.3kg) * (32.85m/s)^2
KE1= 3.65kg * 1079.123m^2/S^2
Ke1=3938.80 joules
Solution
The acceleration of a 7.3 Kg object under the influence of a 32.85 Newton net force will be a=F/m=( 32.85 Newtons)/( 7.3 Kg)= 4.5 meters
per second per second.
In 7.3 seconds the velocity will be 7.3( 4.5 m/s) = 32.85 meters per second.
The average velocity will be the average of this velocity and zero, or ( 32.85 + 0)/2 meters per second = 16.425 meters per second.
At this average velocity, in 7.3 seconds the object will move 119.9025 meters.
The work done will be the product of the distance 119.9025 m and the 7.3 Newton force, or 3938.798 Joules.
Note that the KE of the object at 32.85 m/s will be .5 m v^2 = 3938.798 Joules; thus the work done by the net force is equal to KE gained.
Generalized Solution
If an object of mass m, initially at rest, is acted upon by a net force F for time interval `dt, it will experience acceleration a = F / m
for time interval `dt.
This will result in a velocity change `dv = a `dt = F / m * `dt.
Since the object started from rest its final velocity will be
vf = 0 + `dv = F / m * `dt
and its average velocity will be
vAve = (v0 + vf ) / 2 = (0 + F / m * `dt) / 2 = (1/2) (F / m) `dt.
The distance traveled by the object will be
`ds = vAve `dt = (1/2) (F / m) `dt * `dt = (1/2) (F / m) * `dt^2.
The work done on the object is thus
F * `ds = F * (1/2) ( F / m) * `dt^2 = (1/2) F^2 / m * `dt^2.
The kinetic energy attained by the object will be identical to the work done by the net force:
KE = .5 m vf^2 = .5 m (F / m * `dt) ^ 2 = (1/2) F^2 / m * `dt^2.
Explanation in terms of Figure(s), Extension
The figure below shows the complete relationship between the work done by the net force and the kinetic energy gained by the object.
For the purposes of this problem, note that vf is obtained by the same series of relationships as before and ignore everything to the
right of v0.
Note that if we simply combine our knowledge of vf with the originally given mass m, we obtain KE = .5 m vf^2.
As determined earlier, if v0 = 0 then vf = F / m * `dt, so after a little substitution and simplification we will find that KE = (1/2) F^2
/ m * `dt^2.
Figure(s)
??????Ok, so to find the work done it is best to use 'dw = f * 'ds.....or will this usually be the same??????????????????????????????
The point of this problem is that if you calculate vf and `ds from the given information, and then use vf with the given mass to find KE and `ds with the given force to find `dW_net, you find that `dKE = `dW_net (which is the work-kinetic energy theorem).
This verifies the work-kinetic energy theorem, and the generalized solution shows that it will always be so.
#$&*#$&*
@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
Set 3 Problem number 12
--------------------------------------------------------------------------------
Problem
Solution
Generalized Solution
Explanation in terms of Figure(s), Extension
Figure(s)
Problem
An object with mass 2.1 kilograms is initially at rest.
What will be its KE increase, based on its velocity change, after being acted upon by a net force of 14.7 Newtons for 4.3 seconds?
During the first 4.3 seconds, based on the distance it travels, how much work is done by the net force on the object?
m=2.1kg
v0=0
fnet= 14.7 newtons
'dt= 4.3s
Fnet= a * m
14.7 newtons= 2.1kg * a
a= 7m/s^2
Vf=vo + a * 'dt
Vf= 0 + 7m/s^2 * 4.3s
vf=30.1m/s
'ds= Vave * 'dt
'ds= (30.1m/s + 0m/s)/2 * 4.3s
'ds= 64.72meters
'dw= f * 'ds
'dw= 14.7newtons * 64.72m
'dw= 951.38 joules
KE1= .5m v^2
KE1= .5(2.1kg) * (30.1m/s)^2
KE1= 1.05kg * 906.01m^2/s^2
KE1= 951.31
Solution
The acceleration of a 2.1 Kg object under the influence of a 14.7 Newton force will be a=F/m=( 14.7 Newtons)/( 2.1 Kg)= 7 meters per
second per second.
In 4.3 seconds the velocity will be 4.3( 7 m/s^2) = 30.1 meters per second.
The average velocity will be the average of this velocity and zero, or ( 30.1 + 0)/2 meters per second = `vAve meters per second.
At this average velocity, in 4.3 seconds the object will move 64.71499 meters.
The work done will be the product of the distance 64.71499 meters and the 14.7 Newton force, or 951.3102 Joules.
The kinetic energy will be KE = .5( 2.1 kg)(`speed m/s)^2 = 951.3102 Joules.
Generalized Solution
If an object of mass m, initially at rest, is acted upon by a net force F for time interval `dt, it will experience acceleration a = F / m
for time interval `dt. This will result in a velocity change `dv = a `dt = F / m * `dt. Since the object started from rest its final
velocity will be
vf = 0 + `dv = F / m * `dt
and its average velocity will be
vAve = (v0 + vf ) / 2 = (0 + F / m * `dt) / 2 = (1/2) (F / m) `dt.
The distance traveled by the object will be
`ds = vAve `dt = (1/2) (F / m) `dt * `dt = (1/2) (F / m) * `dt^2.
The work done on the object will be
`dW = F `ds = F (1/2) (F / m) * `dt^2 = (1/2) F^2 / m * `dt^2.
Explanation in terms of Figure(s), Extension
The figure below shows the complete relationship between the work done by the net force and the kinetic energy gained by the object.
If we find `ds as done in previous problems, we can then combine this result with the known force F to obtain the work `dW.
As determined earlier, if v0 = 0 then `ds = (1/2) (F / m) * `dt^2.
After a little substitution and simplification we will find that `dW = F `ds = (1/2) F^2 / m * `dt^2.
Figure(s)
?????I rounded my answeer up a little and it threw the answer for 'dw a little....on these problem should I not round, or is that
diffrence not a big deal????????
A small difference is not big deal, as long as the procedure is correct.
@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
&#Good responses. See my notes and let me know if you have questions. &#