Finding the force of a person given mass.

[mooney82]

Homework Statement

A 30-kg child on roller skates starts up a 10 degree incline at 15 km/h. Assuming she does not propel herself how far up the incline does she travel before stopping? Ignore frictional losses.

Homework Equations

F=ma W=mg

The Attempt at a Solution

I can do this problem if the force is given, but our book doesn't explain how to get force from mass, also would the 15 km/hr have anything to do with it? I've tried to work the problem out several times but I must be using the wrong value for force.

 

[Doc Al]

I can do this problem if the force is given, but our book doesn't explain how to get force from mass,

The force acting is the weight, which you should be able to calculate. What's the component of the weight parallel to the incline?

also would the 15 km/hr have anything to do with it?

The faster her initial speed, the longer it takes her to come to rest.

 

[mooney82]

So the weight is just m*g, which is 294.3 N?

The x component of mg would be 294.3cos10?

 

[Doc Al]

So the weight is just m*g, which is 294.3 N?

Right.

The x component of mg would be 294.3cos10?

No, that's the component perpendicular to the incline (the y-component).

 

[mooney82]

Right.

No, that's the component perpendicular to the incline (the y-component).

But if i set the x-axis on the incline wouldn't the x component be the adjacent side?

 

[Doc Al]

But if i set the x-axis on the incline wouldn't the x component be the adjacent side?

You need to understand how force components are found on inclined planes. Read: http://www.physicsclassroom.com/Class/vectors/u3l3e.cfm"

 

[mooney82]

You need to understand how force components are found on inclined planes. Read: http://www.physicsclassroom.com/Class/vectors/u3l3e.cfm"

Ok, so does the force from the girl start parallel to the ground or the incline? I think that is where I might be getting it wrong. I've been setting it parallel to the ground.

 

[Doc Al]

Ok, so does the force from the girl start parallel to the ground or the incline? I think that is where I might be getting it wrong. I've been setting it parallel to the ground.

The weight acts downward--it has no horizontal component. Find the component parallel to the incline--that's the direction she's moving in.

 

[mooney82]

The weight acts downward--it has no horizontal component. Find the component parallel to the incline--that's the direction she's moving in.

Ahhh I get it, you see the only example my professor gave in class was with an outside force pushing at an object, here there isn't any other force besides her weight, or mg pointing straight down! Thanks, my professor sucks at explaining things, thank you very much!