Nice question!
Use energy.
KE=PE+W
Let x=distance up the slope
(1/2)mv²=mg(sin(&theta))+μmg(cos(&theta))x
Solving for x:
x=(v²/((2g(sin(θ)+μcos(&theta))
A 20 kg box approaches the foot of a 30 degree inclined plane with a speed of 10 m/s. the coefficient of friction between the box and the inclined plane is 0.2. (a) How far up the inclined plane will the box go? (b) If the box slides back down the inclined plane, what is its speed at the bottom of the plane?
2 answers
a. Wb = m*g = 20kg * 9.8N/kg = 196 N. = Wt. of box.
Fp = 196*sin30 = 98 N.=Force parallel
to the incline.
Fn = 196*cos30 = 169.7 N. = Normal or
Force perpendicular to the incline.
Fk = u*Fn = 0.2 * 169.7 = 33.95 N. = Force of kinetic friction.
a = (Fp-Fk)/m = (98-33.95)/20=3.20 m/s^2
V^2 = Vo^2 - 2a*d
d=(V^2-Vo^2)/-2a = (0-100)/-6.4=15.625 m
b. V^2 = Vo^2 + 2a*d
V^2 = 0 + 6.4*15.625 = 100
V = 10 m/s.
Fp = 196*sin30 = 98 N.=Force parallel
to the incline.
Fn = 196*cos30 = 169.7 N. = Normal or
Force perpendicular to the incline.
Fk = u*Fn = 0.2 * 169.7 = 33.95 N. = Force of kinetic friction.
a = (Fp-Fk)/m = (98-33.95)/20=3.20 m/s^2
V^2 = Vo^2 - 2a*d
d=(V^2-Vo^2)/-2a = (0-100)/-6.4=15.625 m
b. V^2 = Vo^2 + 2a*d
V^2 = 0 + 6.4*15.625 = 100
V = 10 m/s.
2 answers