Question

a 10 kg crate slides down a slope that is 3m high and 5 m long with an initial velocity of 4m/s and then slides onto a horizontal surface coming to a stop 5 m after the end of the slope. what is the minimum coefficient of kinetic friction that is required on the horizontal surface to bring the crate to a stop.

Answer 1

vf at top is 4m/s so then
vf2=vi2+2aΛx

so vf at the bottom is 8.65 m/s therefore the deceleration of the object is also
vf2=vi2+2aΛx
and the deceleration is -0.865 if F = m x a then the force acting on the crate is
F = 10 x -0.865
F = -8.65
fk=ukFn

so
uk=???

uk = 0.8826. but this is wrong as the answers i have in the multiple choice give
1)0.3
2)0.32
3)0.6
4)0.66

Answer 2

am i doing this right?

Answer 3

does it say that the slide is frictionless?

Answer 4

yes the only friction is on the bottom horizontal surface

Answer 5

lol i'm stuck too hha

Answer 6

wait 0.76 is the last answer. and then its right. how did you get there? it obviously ran out of typing space and didnt show that last answer.

Answer 7

i hate the fact that these things have a character limit

Answer 8

alright. hold on...

Answer 9

thank you

Answer 10

I'll start with v=8.65 at the end of the slope, what u got was correct.

I have two methods in mind, the easier one being Work=Change in Kinetic Energy. Here is it:

Work done by friction is\[(friction force ) (distance)=-\frac{ 1 }{ 2 }(mass)(8.65m/s)^2\]
\[-(10kg)(9.8m/s^2)(\mu _{k})(5m)=-\frac{ 1 }{ 2 }(10kg)(8.65m/s)^2\] friction force is negative since it's opposite the motion.

Solving, well get uk=0.76..

Answer 11

oh wow thanks man i apreciate it. how is the method i used wrong tho thats what i dont understand.

Answer 12

The second method is using\[\Delta x = \frac{ -v^2 }{ 2a }\] find a and \[a=\frac{ -v^2 }{ \Delta x }\]
Since Force=(Mass)(Acceleration)
\[(mg \mu _{k})=(m)(a)\]
\[\mu _{k}=\frac{ -v^2 }{ g(\Delta x) }\]

Solve and you'll get the same answer. :)

Answer 13

thanks for that one man. I am now certain of what I was uncertain of before! thank you!

Answer 14

You're welcome. I'm glad to help :)

Answer 15

also you wouldnt happen to know what the force responsible for holding a car in a frictionless banked curve is?

Answer 16

is it the horizontal componet of the cars weight?

Answer 17

Hmm.. I'll say it's the normal force the curve exerts on the car.

Answer 18

so its the reaction force to the cars weight?

Answer 19

Yeah.

Answer 20

thanks for that one as well, unfortunatly i cant give u two medals otherwise i would.

Answer 21

The car is moving right? It's also sort of a centripetal force.

Answer 22

yea as it is going around the curve