How to convert 80kgm2sec-1into j-min?

2 Answers
Jan 11, 2018

With dimensional analysis!

#(80kg*m^2)/s * (J)/((kg*m^2)/s^2) * (min)/(60s) approx 1.33J*min#

This makes the case for why units are very important in chemistry and physics.

Jan 11, 2018

#80kgm^2sec^-1 = 1.33 J*"min"#

Explanation:

You have #80kgm^2sec^-1#. The units that they asked you to convert that to, #J*min#, looks like work times time. Allow me to rewrite #80kgm^2s^-1# this way:

#80 kg*(m*m)/s#.

The reason for doing that will become clear.

Newton's 2nd Law, F= m*a, will convince you that the Newton is equivalent to #kg*m/s^2#. Now I will continue to rewrite the original. Now I will rewrite the latest --#80 kg*(m*m)/s#.

#80 kg*(m*m*s)/s^2#

I hope you are following this. That last is an equivalent to the #80kgm^2sec^-1# that you started with.

Now I will group some of the individual units so that I can get Newton*meters, which would be equivalent to Joules. From that last rewrite, check this out:

#80 kg*m/(s^2)*m*s = 80 N*m*s = 80 J*s#

Now we have units of Joule*second. The only thing remaining is to change those seconds to minutes.

#80 J*s * (1 min)/(60 s) = 1.33 J*"min"#

I hope this helps,
Steve