Calculating Power for Pushing a Lawnmower at Constant Speed

  • Thread starter Thread starter ceday
  • Start date Start date
  • Tags Tags
    Power
AI Thread Summary
To calculate the power needed to push a 12 kg lawnmower at a constant speed of 1.2 m/s with a handle tilted at 37 degrees, the gardener must consider the forces acting on the mower, including rolling friction and the normal force. The initial attempt at calculating the force used an incorrect equation, leading to confusion about the normal force and its components. A free-body diagram is recommended to clarify the forces in both the x and y directions. The discussion emphasizes the importance of correctly applying the power equation, which is power equals force times velocity, and ensuring that units are properly understood, as joules measure energy, not power. Further assistance is encouraged by sharing detailed calculations and diagrams.
ceday
Messages
11
Reaction score
0

Homework Statement



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.

Homework Equations



power = force x velocity

The Attempt at a Solution



I know the answer is 24 joules but i can't get there

the equation I used to find the force was F= \mukmg/ \muksin\theta + cos\theta +sin\theta

i know its not right and i can't find the error

thanks!
 
Physics news on Phys.org
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
 
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=-\mukn + fcos\theta where again n= mg + fsin\theta?



I think the term that I am getting hung up on is the normal force
the lawn mower isn't noving up or down so shouldn't the y force = 0?
 
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
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top