Solving the Problem: Calculating Force on a Rock

[29 clicks]

I have this one problem solving question that confuses me. Someone pushes a 50kg rock a distance of 10m in a recent competition of strength. The coefficient of friction between the rock and the ground is 0.8 and the person exerts a force of 400 N. I've got my free body digram with my applied force as the force going foward. I know i have to use Fnet to find the amount of Newtons but what do i use to find fnet. 400N is suppose to minus with something. I'd appreciate it if someone can bring me a bit further into the question so i can get to the answer. I hope this is enough work, I'm trying to get some review before my exams.

 

[andrewchang]

you'll need to subtract the force opposing the push, ie the friction.

 

[29 clicks]

so do i multiply 0.8 (coefficient of friction) by 400 N?

 

[29 clicks]

ok i got 320N as my force of friction (from ff=uFn ff= 0.8x400N)
then i use fnet= fapplied-ffriction to get 400-320=80 N
use fnet in equation fnet=ma
switch it around a= fnet/m a= 80N/50kg which equals 1.6 m/s2
am i right? correct me if I'm wrong please, appreciate it.

 

[29 clicks]

i got the answer now. its 0.16 m/s2 after correcting it.

 

[andrewchang]

the Fnormal would be the force opposing the weight.
which, by the way is equal to mg