# Power lawnmower problem

1. Apr 2, 2009

### ceday

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

A gardener pushes a 12 kg lawnmower whose handle is tilted up 37 deg above horizontal. The lawnmower's coefficient of rolling friction is 0.15 How much power does the gardener have to supply to push the lawnmower at a constant speed of 1.2 m/s Assume his push is parallel to the handle.

2. Relevant equations

power = force x velocity

3. The attempt at a solution

I know the answer is 24 joules but i cant get there

the equation I used to find the force was F= $$\mu$$kmg/ $$\mu$$ksin$$\theta$$ + cos$$\theta$$ +sin$$\theta$$

i know its not right and i cant find the error

thanks!!

2. Apr 2, 2009

### KLoux

Where is your power equation? Show us how you're calculating power and we might be able to help you.

BTW, Joules is a measure of energy, not power... (are you really looking for power?) - In any case, I think you need to show more information (more steps...)

-Kerry

3. Apr 2, 2009

### ceday

Okay

For the Y direction forces, I was using y= -mg+n where n= mg + fsin$$\theta$$?

for x force ,
x=-fk + fcos$$\theta$$ then,
x=-$$\mu$$kn + fcos$$\theta$$ where again n= mg + fsin$$\theta$$?

I think the term that im getting hung up on is the normal force
the lawn mower isnt noving up or down so shouldn't the y force = 0?

4. Apr 2, 2009

### KLoux

Try drawing a free-body diagram of the lawn mower. The forces acting on the lawn mower are allowed to have components in the y-direction. AND what you said is true... the lawn mower is not moving vertically. Is this enough of a hint?

If you still need help with this step, post your free-body diagram (don't forget to include accelerations).

-Kerry