# Work Problem - Pushing a lawnmower

• francisco300
In summary, the lawnmower is pushed a horizontal distance of 10m by a 50 N downward force directed 60 degrees to the horizontal. The coefficient of kinetic frictional force is .017. The work done by each of the external forces is 3.14 N.
francisco300
Work Problem -- Pushing a lawnmower

1. A lawn mower (25kg) is pushed a horizontal distance of 10m by a 50 N downward force directed 60 degrees to the horizontal. The coefficient of kinetic frictional force is .017. What is the work done by each of the external forces on the lawn mower?

## Homework Equations

3. So basically I start off every problem by drawing a free body diagram. I label each force and its direction. I know there is Normal force going up, its Weight going down, and I believe a force going 60 degrees below the horizontal in what would be quadrant 4.

The part I need help in is friction. If you push on a lawn mower the wheels turn clockwise meaning kinetic friction would be pushing right? (This is assuming you picture the lawn mower is being pushed from left to right as is my diagram) I am confused because my instructor did a similar problem where friction was going left which is not opposite of how the wheel turns.

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Where exactly is quadrant 4 ?
If the story says 50 N downward 60 deg wrt horizontal, couldn't it just be someone pushing the handlebar ?
You mention one external force (the 50 N). Why does it ask for work done by each of the external forces ? Any other forces at work here ?

What are the relevant equations you can bring into deal with this problem ?

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• Lawn.jpg
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BvU said:
Where exactly is quadrant 4 ?

BvU said:
If the story says 50 N downward 60 deg wrt horizontal, couldn't it just be someone pushing the handlebar ?
You mention one fecxxternal force (the 50 N). Why does it ask for work done by each of the external forces ?

That is true but usually I like to label forces starting the tail from the center of the object. Is that incorrect? I am not sure what you mean by the last question? I believe because each force can do work depending on the angle?

What are the relevant equations you can bring into deal with this problem ?

Work= F*D*cos θ
Kinetic force= μ*N

Your question is clear now: what about the friction? The exercise is simple: don't bother about the wheels details. There is a weight and an equal and opposite normal force (the mower doesn't move vertically, right?). kinetic friction force you correctly calculate μ*N. Which way is it pointing ? Which way wrt the trajectory ?

Now the importtant thing about work delivered by a force is that the only component of the force that does work is the component along the trajectory (the path). So the friction force is OK. What about me pushing the thing?

Can you calculate the component of the 50 N that does work ?

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You know I can't see your free body diagram, I hope? Very good practice to draw one. From what you mention it looks OK (starting the tail from the center of the object and such) but might not be totally complete.

The relevant equations are OK too and now we have to worry about what to fill in where.

Well I know kinetic friction= μ*N

Normal force≠ Weight in this case because the 50 N at 60° adds a downward force.

So to calculate force I would use this equation:

N-W-Fsin60= 0

I get Normal force = 293.3 N (I use 10 m/s^2 for gravity)

So then Work for Kinetic friction = .017 * 10m * cos θ

Would the angle be 0 or 180?

I sketched my free body diagram in paint. Does it look correct?

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I forgot to welcome you to PF ! Whatever happened to your post #1 ?

Anyway, bedtime for me (GMT+1 = 01:20 AM here.) I have a free body diagram, but I'd rather we use yours.

Do you think you can manage ? You know, like finding the component of the 50 N I'm pushing with that actually does work ?

Some of that work is dissipated (eaten up) by the friction that is pointing against the direction of motion. But I would argue that that is not an external force. Your teacher might disagree, so you be the judge there.

Let me reveal that gravity isn't doing any work: its direction is perpendicular to the motion, so there is no component along the trajectory.

Well I know kinetic friction= μ*N
Normal force≠ Weight in this case because the 50 N at 60° adds a downward force.
So to calculate force I would use this equation:
N-W-Fsin60= 0
I get Normal force = 293.3 N (I use 10 m/s^2 for gravity)
So then Work for Kinetic friction = .017 * 10m * cos θ
Would the angle be 0 or 180?

Again, posts cross, for the umpteenth time tonight. More and more annoying.

You FBD is beautiful.

Friction work: I miss the 293, otherwise: OK.
Angle: I would vote for 180 degrees. 0 degrees is then to the right, where the mower is going.

The other component of the 50 N is more interesting to me: that has to do with the work I am doing.

Thanks for the welcome!

Yeah I can manage the rest of the forces, I just needed help with the work done by frictional force.

The 180 is relevant: the force and the motion are opposite, so it's not doing any work but it is eating it up. Nearly always the case with fricion (there are exceptions, like making wheels turn. That comes later).

I'm almost completely reassured. Just to indulge me: What is the work I have done ?

BvU said:
The 180 is relevant: the force and the motion are opposite, so it's not doing any work but it is eating it up. Nearly always the case with fricion (there are exceptions, like making wheels turn. That comes later).

Oh I see now. I forgot to include the *293.3 in my Work equation but I got the answer correct. Thanks for the help

BvU said:
I'm almost completely reassured. Just to indulge me: What is the work I have done ?

I got 200.2 Joules

Which would me Work done by force which is= 50*cos 60 * 10m = 250

Minus 49.8 J from kinetic friction.

"Answer correct" ? You mean your book tells you -50 J for the friction work ?
Does it also say what the work is that I have done ?

Ah, good we come to the end of this. No adding up of works here. I have done the full 250 J ! No minus anything!

The exercise asks for the work done by each of the external forces. My answer would be 250 J by me and 0 J by gravity (no vertical motion). Also see post #8.

Did you notice some higher authority has edited post #1 to remove the boldface font?

BvU said:
"Answer correct" ? You mean your book tells you -50 J for the friction work ?
Does it also say what the work is that I have done ?

No, one of my classmates got the same answer so i kind of assumed. The question says work done by external forces on the lawn mower. Are the forces I have on my diagram external forces? If so, then what are the internal forces?

Also, is net work the sum of all work no matter if positive or negative?

edit: I did not notice it was edited. Is that bad?

Good questions! Re classmate: if he jumps into the lake, are you going to follow him/her ? (in the latter case i might be mad enough to do so ...;-)

External forces are at least gravity and my pushing. Somewhat debatable what else is external. If the thing stands still, no friction force is present. I would consider that a kind of reaction, but you could also claim that it's the daisies and the grass resisting being pushed down...

Normal force is a reaction force: the Earth holds back the thing from falling in. External ? Doesn't do work anyway: perp to motion.

Nice questions to ask teacher! Earns you credit for being inquisitive and curious. Good teachers like that.

Net sum of work is also sensible to ask for. What does it mean? I push the thing 10 m, costs me 250 J. Friction eats up only 50 J. Where is the 200 ? Did I do -200 J of work to bring it to a halt after 10 m ? Or did I use the brake which dissipated 200 J and changed it into heat to warm up the environment even further ?

More nice questions to ask teacher! I'm useless: falling asleep.

## 1. How does pushing a lawnmower involve work?

Pushing a lawnmower involves work because it requires a force to be exerted on the lawnmower in order to move it. Work is defined as the product of force and displacement, so pushing a lawnmower over a certain distance requires work to be done.

## 2. How does the weight of the lawnmower affect the amount of work required?

The weight of the lawnmower affects the amount of work required to push it because it determines the amount of force needed to move the lawnmower. The heavier the lawnmower, the more force is needed to overcome its weight and push it forward.

## 3. How does the type of terrain affect the work required to push a lawnmower?

The type of terrain can significantly affect the work required to push a lawnmower. Pushing a lawnmower on a flat, even surface requires less work compared to pushing it on a hilly or uneven terrain. This is because the mower needs to be lifted and pushed over obstacles, requiring more force and work to be done.

## 4. How can the angle of the lawnmower's handle affect the work needed to push it?

The angle of the lawnmower's handle can affect the work needed to push it by changing the direction of the applied force. If the handle is tilted upwards, the force applied will have a vertical component in addition to the horizontal component. This means more work is required to push the lawnmower as the vertical component of the force is directed upwards, against the force of gravity.

## 5. How does the condition of the lawnmower affect the work required to push it?

The condition of the lawnmower can affect the work required to push it. A well-maintained and properly lubricated lawnmower will require less work to push compared to a rusty or poorly maintained one. This is because a well-maintained lawnmower will have less friction between its moving parts, making it easier to push.

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