Solving the Problem: Calculating Force on a Rock

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Homework Help Overview

The discussion revolves around a physics problem involving the calculation of force on a rock being pushed across a surface. The scenario includes parameters such as mass, distance, coefficient of friction, and applied force, all of which are relevant to the study of dynamics and frictional forces.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the need to calculate the net force acting on the rock by considering the applied force and the opposing frictional force. Questions arise regarding the correct application of the coefficient of friction and its relationship to the normal force.

Discussion Status

Some participants have made calculations regarding the frictional force and net force, while others are seeking clarification on the steps involved. There is an ongoing exploration of the relationships between the forces, but no consensus has been reached on the final interpretation of the results.

Contextual Notes

Participants are working under the constraints of a homework problem, which may limit the information available for discussion. Assumptions about the system, such as the nature of the forces involved and the conditions of the problem, are being examined.

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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.
 
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you'll need to subtract the force opposing the push, ie the friction.
 
so do i multiply 0.8 (coefficient of friction) by 400 N?
 
Last edited:
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.
 
i got the answer now. its 0.16 m/s2 after correcting it.
 
the Fnormal would be the force opposing the weight.
which, by the way is equal to mg
 

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