# Homework Help: Friction and Forces

1. Nov 26, 2007

### MikalP

1. The problem statement, all variables and given/known data

A 200-kg crate is pushes horizontally with a force of 700 N. If the coefficient of friction is 0.20, calculate the acceleration of the crate.

2. Relevant equations

Well, at the moment I'm just wondering if I used the right equations, and if not then a heads up on where I went wrong

and then proceed to a = Fnet / m?

3. The attempt at a solution

Fn = 200kg x 9.8 = 1960

Ffr = u x Fn, so 1960 x .20 = 392.

Ffr = 392.

a = Fnet / m,

Fnet = Fapplied + Ffr = 700N + 392 = 1092
[Removed Fg and Fn since they cross out as the block is not falling or rising]

Fnet / m = 1092 / 200 = 5.46 m/s

= = = =

I have another question that I'm currently stuck on but have yet to totally give it a try, so I will post the question and update it accordingly.

1. The problem statement, all variables and given/known data

Safety engineers estimate that an elevator can hold 20 persons of 75-kg average mass. The eleveator itself has a mass of 500 kg. Tensile strength tests show that the cable supporting the elevator can tolerate a maximum force of 2.96 x 10^4 N. What is the greatest acceleration the elevator's motor can produce without breaking the cable.

2. Relevant equations

3. The attempt at a solution

2. Nov 26, 2007

### rl.bhat

Fnet = Fapplied + Ffr = 700N + 392 = 1092
Fapplied and Ffr act in the opposite direction. Therefore Fnet = Fapplied - Ffr

3. Nov 26, 2007

### MikalP

Ahh, I always throught it was Fapplied = Fnet - Ffr with Fnet = Fapplied + Ffr.. So it would be 700 - 392 = 308, 308 / 200 = 1.54 m/s. Alright, thanks for the help :) I'll probably come back tomorrow for the next question.