# Newton's Laws and a lawn mower

1. Dec 30, 2009

### seizureboi

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

A person pushes a 14.0-kg lawn mower at constant speed with a force of F= 88.0N directed along the handle, which is set at an angle of 45 degrees to the horizontal. (a) Draw the free-body diagram showing all forces acting on the mower. Calculate (b) the horizontal friction force on the mower, then (c) the normal force exerted vertically upward on the mower by the ground. (d) What force must the person exert on the lawn mower to accelerate it from rest to 1.5 m/s in 2.5 seconds, assuming the same friction force?

2. Relevant equations
m = 14 kg
g = 9.8 m/s2
t = 2.5 seconds
F = ma
NF= mg
v = v0 + at

3. The attempt at a solution

(a) I did the free-body...simple

(b) 88cos(45)-friction force=ma a=0 therefore, 88cos(45)=horizontal friction force= 62N

(c) NF= mg+88sin(45) --> NF= (14)(9.8)+88sin(45)=199.425N

(d) I used the last equation that I put down and i got the acceleration to be .6 m/s2 and I don't know what to do afterwards.

I just think that I'm wrong in every single one...except the free-body =)

2. Dec 30, 2009

### mgb_phys

d, F = m a
You have the 'a' you just need the 'f'

3. Jan 18, 2010

### jeep2717

A car having a mass of 1127 kg is traveling on a level road with a uniform velocity. The horizontal force required to obtain an acceleration of 5 m per sec per sec is?

4. Jan 18, 2010

### jeep2717

If the driver of a pair of gear contains 80 teeth and has a speed of 120 rpm, the speed of the follower gear with 64 teeth is?

5. Jan 31, 2010

### Collisionman

Don't forget the good old

x-x_0=v_0 t+1/2 at^2

... and how to derive it;

x-x_0:distance

average velocity=(v_0+v)/2
average velocity=(x-x_0)/t

(v_0+v)/2=(x-x_0)/t
t(v_0+v)=2(x-x_0)
{v=v_0+at}
x-x_0=1/2 (v_0+v)t
x-x_0=1/2 v_0 t+1/2 vt
x-x_0=1/2 v_0 t+1/2 (v_0+at)t
x-x_0=1/2 v_0 t+1/2 v_0 t+1/2 at^2

x-x_0=v_0 t+1/2 at^2