Find the percentage error in kinetic energy of a body of mass 60.
kinetic energy of a body of mass 60.
0+0.3g and moving with a velocity
of v= 25.0 + 0.cm/s.
Answers
Alright, I’ll express this like this!
Let’s define a couple of variables to account for our error, we’ll call them as follows.
−.3<<.3
a
−
.3
<
a
<
.3
−.1<<.1
b
−
.1
<
b
<
.1
So now, we have the equation for kinetic energy.
=
1
2
2
K
E
=
1
2
m
v
2
Let’s plug in our mass and our velocity.
=
1
2
(60+)(25+
)
2
K
E
=
1
2
(
60
+
a
)
(
25
+
b
)
2
Expand this out and we find…
=
2
2
+25+
625
2
+30
2
+1500+18750
K
E
=
a
b
2
2
+
25
a
b
+
625
a
2
+
30
b
2
+
1500
b
+
18750
Percent error is given by…
=
−1
E
r
r
o
r
P
e
r
c
e
n
t
=
K
E
o
b
v
K
E
t
r
u
e
−
1
Which here would be…
=
2
37500
+
750
+
60
+
2
625
+
2
25
E
r
r
o
r
P
e
r
c
e
n
t
=
a
b
2
37500
+
a
b
750
+
a
60
+
b
2
625
+
2
b
25
You’ll notice that the error percent is only a function of
a
and
b
and that when
a
and
b
are zero (which is to say your error parameter is zero), your error percent is zero.
Now that you have this expression, you can do all kinds of things to it. Maximize and minimize it, figure out what conditions can satisfy a particular error percentage etc.
It could be anything depending on what your errors are (which is random), but now you’ve got the most all encompassing form you can.
If the question was meant to refer to that being the standard deviation of the variable, that’s a bit different.
You can use a bit of a shortcut to get the variance of the function based on the variance of the arguments, which would be…
Take the partial derivative with respect to the original variable using expected value wherever that variable comes up, and multiply that by the squared deviation (variance), and then sum them and take the square root, which here would be…
.3
2
∗
25
2
2
+
.1
2
∗60∗25
‾
‾
‾
‾
‾
‾
‾
‾
‾
‾
‾
‾
‾
‾
‾
‾
‾
‾
‾
‾
‾
‾
√
=6.57
.3
2
∗
25
2
2
+
.1
2
∗
60
∗
25
=
6.57
So your answer could be considered.
18750±.6.57
18750
±
.6
.57
There you go!