Inside a delivery facility, a package (m = 830 kg) is launched with an initial s
ID: 1428310 • Letter: I
Question
Inside a delivery facility, a package (m = 830 kg) is launched with an initial speed of 12.0 m/s at the bottom of a metal ramp (alpha = 28.0 degree). The package and ramp have a coefficient of kinetic friction of 0.330. Do NOT attempt to use conservation of energy to solve this problem. How much time does it take for the package to come to rest (before possibly sliding back down)? Show your work. What maximum vertical height does the package reach? Show your work. What minimum coefficient of static friction would cause the package to remain at rest when it reaches its greatest height? Show your work.Explanation / Answer
given
m = 8.5 kg
initial speed, u = 12 m/s
alfa = 28 degrees
mue_k = 0.33
a) Net force acting on the package, Fnet = -m*g*sin(28) - mue_k*N
m*a = -m*g*sin(28) - mue_k*m*g*cos(28)
a = -g*sin(28) - mue_k*g*cos(28)
= -9.8*sin(28) - 0.33*9.8*cos(28)
= -7.46 m/s^2
final speed, v = 0
let t is the time taken
Apply, v = u + a*t
==> t = (v - u)/a
= (0 - 12)/(-7.46)
= 1.608 s
b) distance travelled along the path, d = (v^2 - u^2)/(2*a)
= (0^2 - 12^2)/(2*(-7.46))
= 9.65 m
so, maximum vertical height reached, hmax = d*sin(28)
= 9.65*sin(28)
= 4.53 m
c) when Fnet = 0, on the packege at maximum height it will not slight back.
m*g*sin(28) - mue_s*N = 0
m*g*sin(28) - mue_s*m*g*cos(28) = 0
m*g*sin(28) = mue_s*m*g*cos(28)
==> mue_s = tan(28)
= 0.5317