Question

A crate slides down an incline plane and reaches the horizontal surface with a speed of 5m/s. it slides for 2m along this surface, with a coefficient of kinetic friction of 0.3. what is the velocity of the crate at the end of the 2m?

Answer 1

first, write your given
Vi=initial velocity=5m/s
d=2m
coefficient of friction = 0.3

analysis:
you know that
Fnet=ma,
and Fk= μk•Fnormal
Fnormal= mg
Fnet=Fk
Fnet=μk•Fnormal
ma=μk•mg

using the 5 key equations, \(v_f^2=v_i^2 +2ad\)
solve for a
\(a=\frac{v_f^2-v_i^2}{2d}\)
substitute from above equation:
\(m(\frac{v_f^2-v_i^2}{2d})=μ_kmg\)
then we know that m will gonna cancel out, so it will be:
\((\frac{v_f^2-v_i^2}{2d})=μ_kg\)
then evaluate this again to solve for \(v_f\)
and done!

Answer 2

Am i correct with Vf = 2.352 m/s?

Answer 3

hmm.. i got 6.06 m/s

Answer 4

F=ma
Force gravity-Force friction = ma
W-Fk=ma hence, Fk=Uk(N)
mg-Uk(mg)=ma
g-Uk(g)=a
a=9.81-0.3(9.81)
a=6.87 m/s^2

To find the Velocity (final) with respect to the distance on the surface:
Vf^2=Vi^2+2a(delta x) where it is noted, that we use Newton Second Law of Kinematic Equation.
a=(Vf^2-Vi^2)/2(delta x)
6.87m/s^2=[Vf^2-(5)^2] / 4m
Vf^2=52.48
Vf=7.24 m/s

Answer 5

Did you learn conservation of energy?

Answer 6

mass doesn't enter into equation. The crate would slide the same if it had mass of 1000kg or one kg. More mass makes more normal force but more mass gives more momentum, so the masses cancel in effect:

(Vf^2-Vi^2)/2(delta d) = u * g

d = V^2 / 2* u *g
d = distance in meters
V f= final velocity of crate
Vi = initial velocity of crate
u = coefficient of kinetic friction = 0.30
g = gravity acceleration = 9.81 m/s^2

1/4 (Vf^2 - 25) = 2.943
solve for V^2:
Vf = 6.06 m/s