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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 belt

Attempted 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/s

this 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 belt

Attempted 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/s

this 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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