Question #3df20

1 Answer
Jan 20, 2017

See below.

Explanation:

The weight of a stationary object near the surface of the earth is equal the product of the mass of the object and the gravitational (free-fall) acceleration constant, #g#. This is equal in magnitude to the force of gravity.

#W=vecF_g=mg#

Where #W# is the weight.

Weight is a force and can be given in newtons, where a newton is equivalent to one kilogram meter per square second:

#1 N= (1kgm)/s^2#

If using newtons, the mass of the object should be in kilograms. The gravitational acceleration constant, #g#, is equal to #9.8m/s^2#. So, if we are calculating the weight of a stationary object on Earth we would have units:

#W=mg=kg*(m/s^2)#

#=>(kgm)/s^2#

So, you should get units of newtons, #N#.

Multiplying #kg*N/(kg)# would get you the correct units. The difference is that this expresses the units of acceleration as #N/(kg)#, which is equivalent to #((kgm)/s^2)/(kg)=m/s^2#. It is not incorrect.