Calculating Forces on Inclined Surfaces

[sbot]

Homework Statement

1) The coefficient of kinetic friction for a 22-kg bobsled on a track is 0.10. What force is required to push it down at a 6.0 degree incline and achieve a speed of 60 km/h at the end of 75 m?

2) A box is given a push so that it slides across the floor. How far will it go, given that the coefficient of kinetic friction is 0.20 and the push imparts an initial speed of 4.0 m/s?

Homework Equations

F = ma
Force of friction = (mu)k x normal force

The Attempt at a Solution

I calculated the net force in #1 to be 1.06 after getting the components of gravity on the box, but this seems incorrect. I don't know where to go for both problems, especially #2. Where do you start when the mass of the box is not given?

 

[pooface]

I would like to learn how to solve this problem as well as I am learning this myself.
I find that:
Fw = 22kg*g = 215.82N
FN = 215.82Ncos6deg = 214.64N
------ these two values confirm that the coeff of kinetic friction is 0.10
kinetic friction force = (mu)k*FN = 21.464N
using F=ma, F/m = a (unsure about this step)
21.464N/22kg = 0.9756 m/s^2
------At 75m we want a speed of 60 km/h.
dont know what to do from here. I actually have to study for another test so I will try later.

 

[ytoruno]

well for problem 2, you ask about the lack of a given mass. Don't worry about such things, if there's no mass value just use m, usually such simple problems yeild differential equations in which the value m cancels out.

heres a detailed solution for 2):
first make a sum of the forces in one direction : ma=mx''= -(uk)mg {now remember that friction is a resistive force, thus the minus sign}

so x''=a= -(uk)g = -1.96 m/s^2 but this can also be seen as a differential equation

x''= dv/dt and the equation is seperable

thus dv=-1.96 dt and we integrate accordingly

v= -1.96(t) + C with the initial condition V(0)= 4m/s

V(0)= 4 = C

the we have solved v(t)= -1.96(t) + 4

to find out when it stops we will solve for when velocity is zero

-1.96(t)+4=0 ==> t=2.04 seconds is the time it stops after the initial push

now to find the position, we integrate v(t)=dx/dt further

x= (1/2)(-1.96)(t^2) +4t + C and we can impose the condition that the box was at the origin at the beggining, thus x(0)= 0

yields x(t)=(1/2)(-1.96)(t^2) +4t

and if we plug in the stop time, you will get the final position.

 

[pooface]

what about number one? can you give some help?

 

[ytoruno]

problem 1 is only slightly more complex
the sum of force F=ma=mx'' is:

mx''=-ukmgcos(s)+mgsin(s) where a = the angle 6.0 degrees

then x'' = g[sin(s)-ukcos(s)] = .049 m/s squared, we surmise at this point that the intial velocity has to be very close to the final velocity since this acceleration is small.

x'' =dv/dt= .049 which yields v=.049t + C and we impose V(0)=Vi (initial velocity)

so v(t)= .049t +Vi

integrating further we find

x(t)= .0245(t^2)+Vi(t)+ B where B will be zero since we impose x(0)= 0
we set x= 75 meters solve for t with the quadratic formula

t= [-Vi+sqrt(Vi^2 - .098(-75))] / .049

we plug this into the t of the v(t) equation and solve for Vi

Vi= 59.8 m/s , a value very close to 60m/s due to the very low angle and frictional resistance.

 

[sbot]

thankyou ytoruno that helps a lot