Question

The masses of the objects are m1 = 17.0 kg and m2 = 10.5 kg, the mass of the pulley is M = 5.00 kg, and the radius of the pulley is R = 0.300 m. Object m2 is initially on the floor, and object m1 is initially 4.40 m above the floor when it is released from rest. The pulley's axis has negligible friction. The mass of the cord is small enough to be ignored, and the cord does not slip on the pulley, nor does it stretch.
How much time (in s) does it take object m1 to hit the floor after being released?

Answer 1

Answer:

Explanation:

Given:

m1 = 17 kg

m2 = 10.5 kg

M = 5 kg

r = 0.3 m

Use the energy equation

Initial Energy = Final Energy

Initial Energy is only the potential energy of mass one (PE1).

Final Energy is the final kinetic energy of mass one (KE1), the final kinetic energy of mass two (KE2), the kinetic energy of the pulley (KEp) and the potential energy of mass two (PE2)

PE1 = KE1 + KE2 + KEp + PE2

PE1 = m1*g*h

PE2 = m2*g*h

KE1 = (1/2) m1 * v^2

KE2 = (1/2) m2 * v^2

KEp = (1/2) I w^2

I = (1/2) M r^2

w = (v/r)

PE1 = KE1 + KE2 + KEp + PE2

m1*g*h = (1/2) m1 * v^2 + (1/2) m2 * v^2 + (1/2) I w^2 + m2*g*h

Now substitute in I = (1/2) M r^2 and w = (v/r) inti the above equation, we have:

m1*g*h = (1/2) m1 * v^2 + (1/2) m2 * v^2 + (1/2) (1/2) M r^2 (v/r)^2 + m2*g*h

m1*g*h = (1/2) m1 * v^2 + (1/2) m2 * v^2 + (1/4) M v^2 + m2*g*h

m1*g*h - m2*g*h = v^2 [(1/2) m1 + (1/2) m2 + (1/4) M]

v^2 = (m1*g*h - m2*g*h)/[(1/2) m1 + (1/2) m2 + (1/4) M]

v^2 = g*h*(m1 - m2)/[(1/2) m1 + (1/2) m2 + (1/4) M]

v = sqrt[g*h*(m1 - m2)/[(1/2) m1 + (1/2) m2 + (1/4) M]]

Inputting values,

v = sqrt[9.81 * 4.40 *(17 - 10.5)/[(1/2) × 17 + (1/2) × 10.5 + (1/4) × 5]]

v = 4.325 m/s

There is one motion equation without acceleration

H = S + (1/2) * (vi - v0) * t

S = 0 m

v0 = 0 m/s

h = vi*t/2

t = 2 * H/vi

= 2 * 4.40/4.325

= 2.035 s.