Solving Force and Rest with a Skater on Ice

[enantiomer1]

Homework Statement

here's the problem
A 68.5 kg skater moving initially at 2.40 m/s on rough horizontal ice comes to rest uniformly in 3.52 s due to friction from the ice.

Homework Equations

F= ma

The Attempt at a Solution

I'm sure this equation is reliant on finding the acceleration, but I really don't know how to do that when time (instead of distance) is our given variable
anyone know what I'm missing?

 

[enantiomer1]

okay sorry scratch that I just had a brain fart or something -.-
obviously a= m/s² so there for all I had to do was divide the velocity by the time
I swear to god I'm getting stupider sometimes

 

[thomate1]

Your approach to the problem is correct. In order to solve for the force of friction, we need to first find the acceleration of the skater. This can be done by using the equation F = ma, where F is the force of friction, m is the mass of the skater, and a is the acceleration.

To find the acceleration, we can use the equation a = (vf - vi)/t, where vf is the final velocity (in this case, 0 m/s since the skater comes to rest) and vi is the initial velocity (2.40 m/s). Substituting in the given values, we get a = (0 - 2.40 m/s)/3.52 s = -0.68 m/s^2.

Now, we can use the equation F = ma to find the force of friction. Substituting in the mass of the skater (68.5 kg) and the acceleration we just calculated (-0.68 m/s^2), we get F = (68.5 kg)(-0.68 m/s^2) = -46.58 N.

Therefore, the force of friction acting on the skater is -46.58 N, which is the force that causes the skater to come to rest in 3.52 s.