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pleasehelpme6
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Packages are thrown down an incline at A with a velocity of 1 m/s.
The packages slide along the surface ABS to a conveyor belt which moves with a velocity of 2 m/s.
Knowing that d = 7.5 m and µ = 0.25 between the packages and all surfaces, determine...
a) speed of the package at C
b) distance a package will slide on the conveyor belt before it comes to rest relative to the beltAttempted solution:
FBD -- N = Wsinθ for part 1
Kinetic energy = T
Work = U
T2 = T1 + U12
T2 = 1/2 mv^2 + [Fweight - Ffriction]*distance
T2 = m*[1/2+ Wsinθ-µWcosθ]*(7.5 m)
T2 = 21.36*m (Energy at point B)Then...
N = W = mg
T3 = T2 + U23 = T2 -(Ffriction*distance)
T3 = 21.36*m - .25*mg*7
T3 = 4.19*m
4.19*m = 1/2 mv^2
v = sqrt(4.19*2)
v=2.89 m/sthis is the correct velocity at the point, but when i try to find the distance it will slide, i cant.TF= Tfinal = 0
Tf = T3 + U3->F = 0
T3 = -U3F
4.19*m = µ*mg*distance
distance = 4.19/(µ*g)
this gives me a distance of 1.7 m, but the answer is 0.893 m.
help please? thanks
The packages slide along the surface ABS to a conveyor belt which moves with a velocity of 2 m/s.
Knowing that d = 7.5 m and µ = 0.25 between the packages and all surfaces, determine...
a) speed of the package at C
b) distance a package will slide on the conveyor belt before it comes to rest relative to the beltAttempted solution:
FBD -- N = Wsinθ for part 1
Kinetic energy = T
Work = U
T2 = T1 + U12
T2 = 1/2 mv^2 + [Fweight - Ffriction]*distance
T2 = m*[1/2+ Wsinθ-µWcosθ]*(7.5 m)
T2 = 21.36*m (Energy at point B)Then...
N = W = mg
T3 = T2 + U23 = T2 -(Ffriction*distance)
T3 = 21.36*m - .25*mg*7
T3 = 4.19*m
4.19*m = 1/2 mv^2
v = sqrt(4.19*2)
v=2.89 m/sthis is the correct velocity at the point, but when i try to find the distance it will slide, i cant.TF= Tfinal = 0
Tf = T3 + U3->F = 0
T3 = -U3F
4.19*m = µ*mg*distance
distance = 4.19/(µ*g)
this gives me a distance of 1.7 m, but the answer is 0.893 m.
help please? thanks
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