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Vinil7 [7]
3 years ago
7

A hollow cylinder of mass 2.00 kg, inner radius 0.100 m, and outer radius 0.200 m is free to rotate without friction around a ho

rizontal shaft of radius 0.100 m along the axis of the cylinder. You wrap a light, nonstretching cable around the cylinder and tie the free end to a 0.500 kg block of cheese. You release the cheese from rest a distance h above the floor. If the cheese is moving downward at 4.00 m/s just before it hits the ground, what is the value of h?
Physics
1 answer:
Aneli [31]3 years ago
5 0

Answer:

h=2.86m

Explanation:

In order to give a quick response to this exercise we will use the equations of conservation of kinetic and potential energy, the equation is given by,

\Delta PE_i + \Delta KE_i = \Delta PE_f +\Delta KE_f

There is no kinetic energy in the initial state, nor potential energy in the end,

mgh+0=0+KE_f

In the final kinetic energy, the energy contributed by the Inertia must be considered, as well,

mgh = (\frac{1}{2}mv^2+\frac{1}{2}I\omega^2)

The inertia of the bodies is given by the equation,

I=\frac{m(R_1^2+R^2_2)}{2}

I=\frac{2(0.2^2+0.1^2)}{2}

I=0.05Kgm^2

On the other hand the angular velocity is given by

\omega =\frac{v}{R_2}=\frac{4}{1/5} = 2rad/s

Replacing these values in the equation,

(0.5)(9.8)(h) =\frac{1}{2}*0.5*4^2+\frac{1}{2}*0.05*20^2

Solving for h,

h=2.86m

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v_f = 61 km/h = 16.9 m/s
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(b) The average power delivered by the engine of the car during the 19 s is equal to the work done by the engine divided by the time interval:
P= \frac{W}{\Delta t}
But the work done is equal to the increase in kinetic energy of the car, and since its initial kinetic energy is zero (because the car starts from rest), this translates into
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(c) The instantaneous power is given by
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So we need to find the acceleration first:
a= \frac{v_f-v_i}{\Delta t}=  \frac{16.9 m/s}{19 s}=0.89 m/s^2
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ratelena [41]

Answer:

Explanation:

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Answer:

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Explanation:

Given data:

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