# Error Propagation in Measuring the Kinetic Energy

For a function of several variables, $f = f(a,b,c,…)$, the propagation error in a measurement of $f$ is

$\sigma^2_f = (\frac{\partial f}{\partial a})^2 \sigma^2_a + (\frac{\partial f}{\partial b})^2 \sigma^2_b + (\frac{\partial f}{\partial c})^2 \sigma^2_c +…$

In your next lab, you will be introduced to the concept of kinetic energy,

$K = (1/2)m v^2$ where $m$ is the mass of the object and $v$ is its speed having the following values:

$m = 4.7 \pm 0.5 (kg)$

$v = 0.4 \pm 0.2 (m/sec)$

Using the method of propagation errors, what is the kinetic energy and its uncertainty? Make certain you carefully show all your work.

The total kinetic energy is

$K = (m*v^2)/2 = (4.7*0.4^2)/2 =0.376 Joules$

Differentiating twice the equation of Kinetic energy we obtain

$d K = (d K/d m)*d m + (d K/d v)*d v$

$d K^2 = (d K/d m)^2*(d m^2) + (d K/d v)^2*(d v^2)$

We know that

$(d K/d m) = (1/2)v^2 =0.5*0.4 =0.2$

and

$(d K/d v) = m v = 4.7*0.4 =1.88$

with the errors in mass and speed given as

$d m =0.5 kg$ and $d v =0.2 m/s$

Therefore

$(d k)^2 = 0.2^2 * d m^2 +1.88^2*d v^2 =0.04*0.25 +3.5344*0.04 =0.151376$

Dave

February 9, 2022 @ 10:06 pm

Computing (dK/dm)=(1/2)v^2 = 0.5*0.4=0.2 is not correct. It should be (1/2)v^2 = 0.5*0.4*0.4 = 0.08.