Why is acceleration inversely proportional to mass?

2 Answers
Apr 28, 2018

Answer:

acceleration equals to the force applied divided by mass

Explanation:

an object moving at a velocity of x carries the force of its mass times its speed.

when you apply a force onto an object, the increase in speed of it would be affected by its mass. Think of it this way: you apply some force onto an iron ball, and apply the same force on a plastic ball (they are of equal volume). Which one moves faster, and which one moves slower? The answer is obvious: the iron ball will accelerate slower and travel slower, while the plastic ball is faster.

The iron ball has a greater mass, so the force which makes it accelerate is deduced more. The plastic ball has a smaller mass, so the force applied is divided by a smaller number.

I hope this helps you a bit.

Apr 28, 2018

Answer:

Assuming we're using #F=ma#, then it's because, when one goes up, the other must go down in order to keep the equation balanced.

Explanation:

Say we wish to keep a force #F# exerted by an object constant. If the mass #m# of the object doubles, what must happen to the object's acceleration #a# to keep #F# unchanged?

The answer is: the object's acceleration must be halved.

We start with

#F=m*a#

and if we double the mass to #2m#, the RHS as a whole has doubled. Thus, the LHS also doubles, meaning we get double the force:

#2F = 2m*a#

This is an example of direct proportionality between #F# and #m#. If #m# doubles, #F# responds by doubling as well.

But we want to keep the force the same; we don't want #2F#, we want #F#. So we need to divide the LHS by 2. And to do that, we must divide the RHS by 2 as well. So either the mass #2m# goes back down to #m#, or the acceleration #a# gets cut to #1/2 a#.

#F= 2m*1/2 a#

This is an example of inverse proportionality. When the force is taken as a constant, if mass doubles, acceleration must be halved.

Note:

You can also see the inverse relation between #m# and #a# by solving #F=ma# for one or the other.

#F=ma " "=>" "a=F/m" " <=>" "a=F(m^-1)#

#color(white)(F=ma) " "=>" "m=F/a" "<=>" "m=F(a^-1)#

It's now easy to see mathematically that #a# and #m# are inversely proportional, because each is a multiple of the other's inverse (that multiple being #F# itself).