A net force of 10N acts on a mass of 25kg for 5 seconds. What is the acceleration?

I can't use #F=ma#, because it doesn't account for time. I'm stuck.


2 Answers
Apr 28, 2018

The acceleration is going to be zero, assuming that the mass is not sitting on a frictionless surface. Does the problem specify a coefficient of friction?


The 25 kg object is going to be pulled down on whatever it is sitting on by the acceleration due to gravity, which is approx
#9.8 m/s^2#.

So, that gives 245 Newtons of downward force (offset by an upward normal force of 245 Newtons provided by the surface it's sitting on).

So, any horizontal force is going to have to overcome that 245N downward force (assuming a reasonable coefficient of friction) before the object will move.

In this case, the 10N force will not be enough to make it move.

Apr 28, 2018

#a = 0.4 m/s^2#


That 5 seconds is thrown into the question to see if you can be confused by extraneous information.

You have net force and mass, therefore you can use #F_"net" = m*a#.

#F_"net" = 10 N = 25 kg*a#

Solving for a,

#a = (10 N)/(25 kg) = 0.4 m/s^2#

That value of acceleration was the acceleration for the entire time the force was applied.

I hope this helps,