Stopping distance and forces!

ok so i am studying for finals...i mean who isnt and i have a couple questions....i got them wrong previously so maybe ya'll can help me firgure them out...ok

sally is riding her bicycle on a path when she comes around a corner and sees that a fallen tree is blocking the street 42 meters ahead. if the coefficient of friction between her bike's tires and the gravel path is .36 and she is traveling at 50000 meters per hour how much stopping distance will she require? sally and her bike have a mass together of 95 kilograms and also

a box weighing 215 newtons is placed on an inclined plane that makes a 45.0 degree angle with the horizontal...compute the component of the gravitational force acting down the inclined plane....if the coefficient of static friction has a maximum value of .35, will the box slide down the plane?

i can draw the pictures for both i just do not know what to do next...please any help you have would be great
 

lightgrav

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You've done work on these already? what did you do?
I hope you converted the 50 km/hr into m/s ...
 
what are you talking about there is no 50 km/hr.....but i did convert the 50000 meters per hour to meters per second and i got 800 meters per second and then that would be initial velocity and then force friction equals mass times gravity right? but then i have initial and final velocity, the mass, distance to the fallen tree and coefficient of friction so how can i solve for stopping distance?
 

lightgrav

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50,000 meters is 50 km. get distance from acceleration!

Friction Force would have to include the co-efficient of friction ...
What's the total horizontal Force?
You don't know the mass of the moving object; call it "m", it will cancel.

Generally, once you've drawn the Force diagram,
compute the Force components parallel to the velocity ...
the Sum of Force components equals ma .
 
sorry! ok so anyways 800 meters per second is initial velocity and wait the mass is 95 kilograms... so you have coefficient of friction = gravity divided by acceleration? so then you just have gravity divided by the coefficient so then acceleration = 27.2 meters per second squared and then how does that help me? i think i just did stuff incorrectly or stuff i didnt need to do....thanx for all your past and future help...
 

lightgrav

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[tex]F_{friction} = \mu * F_{Normal} = \mu m g[/tex]

which is what causes the acceleration (ma) .

Once you get the acceleration, you can use
[tex]v_f^2 = v_i^2 + 2 a x [/tex]
 
ok so Force friction is equal to -340 Newtons divided by the mass of 95 kg would make the acceleration -3.6 meters per second squared....but then my initial velocity has to be wrong because velocity final is 0 but if i input all of my values 800 meters per second for initial velocity and -3.6 meters per second squared as acceleration then x would equal 89000 m which is no way possible and then i have a question cuz i never used the distance between sally and the tree (42 meters)
 

lightgrav

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You only check whether the distance needed to stop is beyond the tree ...
if the bike hits the tree, the Force by Tree will stop the bike there.

do you know how fast 800 m/s is? 800 m is about half a mile ... in 1 sec ... Vance would be jealous! Want to show me how you converted?

50,000 m/hr =
 
ok i said (50,000m/hr)(1hr/60s) = 800 m/s
 

lightgrav

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First, don't round off so quickly. keep 3 or 4 digits during calculations,
or round-off errors will accumulate horribly!

second, if you wait 60 seconds, how long is that?
 
dang it!!! i am retarded...there are 3600s in an hour so then after all the calculations my final answer is 27.3m....thank you for all your help you really stuck with me!!!!
 

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