Question 14 of 16 Mapoob Sapling Learning Two buckets of mass m-20.3 kg and m2 1

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Question 14 of 16 Mapoob Sapling Learning Two buckets of mass m-20.3 kg and m2 15.1 kg are attached to the ends of a massless rope, which passes over a pulley with a mass of mp 7.73 kg and a radius of rp 0.150 m. Assume that the rope does not slip on the pulley, and that the pulley rotates without friction. The buckets are released from rest, and begin to move. If the larger bucket is a distance do-2.05 m above the ground when it is released, with what speed v will it hit the ground? Number m/s 20.3 kg 15.1 kg Hint Previous Give Up & View Solution Check Answer Next Exit

Explanation / Answer

You need to take the initial energies and set them equal to the final energies of each mass including the pulley using KE=1/2mv^2 PE=mgh and KE=1/2Iw^2

The initial energy would just be the potential energy of the first mass making it:
m1*g*h or m1*g*d in this case

The final energy is a bit more difficult because you have to remember that m2 has both types of energies because the weight will end up where the first weight started, so the equation would look like:

(m2*g*d)+(1/2*m1*v^2)+(1/2*m2*v^2)+(1/2...

now I = 1/2mr^2 because it is treated as a solid disk and we know that w= v/r from formulas

plugging that in will look like (1/2*1/2*mp*r^2*v^2/r^2) and the r^2 would cancel making it 1/4*mp*v^2

setting the initial and final equations equal would give you:

m1*g*d = (m2*g*d)+(1/2*m1*v^2)+(1/2*m2*v^2)+(1/4*...

now we want to simplify and solve for v, and by using basic algebra you come up with v

(m1*g*d) - (m2*g*d) = (1/2*m1*v^2)+(1/2*m2*v^2)+(1/4*mp*v^2)
(m1*g*d) - (m2*g*d) = v^2((1/2*m1)+(1/2*m2)+(1/4*mp))
((m1*g*d) - (m2*g*d))/ ((1/2*m1)+(1/2*m2)+(1/4*mp)) = v^2

v= sqrt(((m1*g*d) - (m2*g*d))/ ((1/2*m1)+(1/2*m2)+(1/4*mp)))

v= sqrt(((20.3*9.81*2.05) - (15.1*9.81*2.05))/ ((1/2*20.3)+(1/2*15.1)+(1/4*7.73)))
v = 2.307m/s

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