Lawn mower force question, and acceleration.

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A person pushes a 16 kg lawn mower at a constant speed with a force of 80 N directed along the handle which is set at an angle of 45 degrees to the horizontal. If the coefficient of friction against the lawn mower is 0.230, what force must the person exert on the lawn mower to accelerate it from rest to 1.5 m/s over 3.0 m??

so i put the 80 N into a right triangle as the hypotenuse, and solved for the x and y terms and got 57 N. I then put it into a freebody diagram and and now stumped because i don't know what to solver for. If anyone could help its greatly appreciated! thanks!:smile:
 
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so to find the acceleration i would use v1= 0, v2 = 1.5 and Delta displacement is 3.0 m and use a kinematic then how would i find the net force?
 
bobsagget said:
so to find the acceleration i would use v1= 0, v2 = 1.5 and Delta displacement is 3.0 m and use a kinematic

Yes :smile:
… then how would i find the net force?

oh come on! :rolleyes:

what is the relationship between net force and acceleration??
 
hahaha wow I am slow, So i use the F=ma formula which would give me a net force of 6N. so then i have to use fnet= force applied-force of friction, and i would want to solve for the for F(app) using a free body diagram.
 
and i also solved for net force and i got 29 N. so would i now need to put that into a right triangle with a 45 degrees angle and the 29 N would be on the x-axis of that triangle? and id need to solve for the hyp? then do i add it to the 80 N originally to find the total force that is needed to accelerate the lawn mower?
 
bobsagget said:
hahaha wow I am slow, So i use the F=ma formula which would give me a net force of 6N. so then i have to use fnet= force applied-force of friction, and i would want to solve for the for F(app) using a free body diagram.

Yes!

Now do it!​