What are the units of #k# in the expression #F=kv^n# (a) when #n=1# and (b) when #n=2#?

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Answer 1 · 2017-07-09T03:27:59

(a) When #n=1# the units of #k# are #Nm^-1s#.

(b) When #n=2# the units of #k# are #Nm^-2s^2#

We can rearrange the equation to make #k# the subject, since that is what we are interested in:

#k=F/v^n#

If #n=1# this becomes simply:

#k=F/v#

In terms of units, this is #N/(ms^-1)#, which we can write as #Nm^-1s# (because #1/s^-1=s#).

When #n=2# we have:

#k=F/v^2#

The units will be #N/(ms^-1)^2= N/(m^2s^-2)#, which we can write as #Nm^-2s^2#.