A conveyor belt system delivers each 10 kg crate filled with plastic dolls to the ramp  at point A, such that the crate's velocity is 2.0 m/s, directed down the ramp. If the coefficient of kinetic friction between the crate and the ramp is 0.3, determine the speed at which each crate slides off the ramp at B and onto the warehouse cart.

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The figure is attached below.

On the ramp the weight of the crate decomposes along the ramp and normal to the ramp (`alpha=30 deg` is the angle of the ramp)

`G_p =G*sin(alpha)`

`G_n = G*cos(alpha)`

On the axis parallel to the ramp we have

`G_p -F_f = m*a`

`G_p -mu*G_n...

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The figure is attached below.

On the ramp the weight of the crate decomposes along the ramp and normal to the ramp (`alpha=30 deg` is the angle of the ramp)

`G_p =G*sin(alpha)`

`G_n = G*cos(alpha)`

On the axis parallel to the ramp we have

`G_p -F_f = m*a`

`G_p -mu*G_n =m*a`

`m*g*sin(alpha) -mu*m*g*cos(alpha) =m*a`

`a =g*[sin(alpha)-mu*cos(alpha)] =9.81*[sin(30) -0.3*cos(30)]=2.36 m/s^2`

Now, the length of the ramp is given in figure. The relation between acceleration, displacement and initial end final speed is (for uniform accelerated motion)

`v_B^2 =v_A^2+2*a*L`

`v_B=sqrt(2^2 +2*2.36*10) =sqrt(51.2) =7.15 m/s `

Answer: the speed of crate along the ramp at point B is 7.16 m/s

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