I need help with the following problem, I tried doing it so many times but I get stuck.
A 4 kg sleigh has an initial speed of 10 m/s in the peak of a 34° slope. If uk = 0.2, ¿what distance will the sled travel when its speed reaches 30 m/s?
What I tried to do was:
[ (4 kg)(9.8 m/s)(s * sin 34) + [(4 kg)(10 m/s)^2]/2 ] = [ [(4 kg)(30 m/s)^2]/2 + (fk * s) ]
I got fk with =
Fy = 0
Fn = 39.2 N (Sin 56)
Fn = 32.49 N
fk = MkFn
fk = (0.2) (32.49 N)
fk = 6.49 N
I really need a hand on this one. Thanks in advance.
2007-03-07 17:50:55 · 3 answers · asked by ddeity_inc 3 in Science & Mathematics ➔ Physics
The answer is 104 m but I don't know how to get to the answer.
2007-03-07 17:52:42 · update #1
you're right, the answer is 103,9,,you can use this formula:
s=(V^2-Vo^2)/(2*a) and V=30m/s , Vo=10m/s , and you can get the acceleration a=((mg*sin34)-0,2*(mg*cos34))/4,,, and that's it!!!
2007-03-07 18:10:24 · answer #1 · answered by ivandavid 1 · 0⤊ 1⤋
I am not a mathematician. Where in your equation is the original 10 m/s being changed from. I don't understand why it is not constant. How long does the acceleration from 10m/s to 30m/s (you know the 20m/s gain) take and how far is the distance in the original 10m/s for that time? I think it would be simpler two treat it as to separate problems.
then just add them together.
How much distance to get to 20m/s + 10m/s times how long it takes to accelerate from 0 to 20 m/s
Really just guessing
I am not real smart about this.
2007-03-08 02:03:13 · answer #2 · answered by U-98 6 · 0⤊ 1⤋
looks like it's 103.7 m
acceleration along the slope = gravitational component - frictional component
=g*(sin(theta) - uk*cos(theta)
delta_v = acceleration*time
distance = v0*time+1/2acceleration*time^2
2007-03-08 02:13:36 · answer #3 · answered by arbiter007 6 · 0⤊ 0⤋