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Nookie1986 [14]
3 years ago
8

A box of unknown mass is sliding with an initial speed vi = 5.60 m/s across a horizontal frictionless warehouse floor when it en

counters a rough section of flooring d = 2.30 m long. the coefficient of kinetic friction between the rough section of flooring and the box is 0.100. using energy considerations, determine the final speed of the box (in m/s) after sliding across the rough section of flooring.
Physics
1 answer:
Maksim231197 [3]3 years ago
6 0

We know that the Delta E + W(Work done by non-conservative forces) = 0 (change of energy)

In here, the non-conservative force is the friction force where f = uN (u =kinetic friction coefficient) 

W= f x d = uNd ; N=mg 
Delta E = 1/2 mV^2 -1/2mVi^2 

umgd + 1/2mV^2 - 1/2mVi^2 = 0 (cancel out the m term) 

This will then give us: 

1/2Vi^2-ugd = 1/2V^2 

V^2 = Vi^2 - 2ugd

So plugging in our values, will give us:

V= Sqrt (5.6^2 -2.3^2)

=sqrt (26.07)

= 5.11 m/s 

 

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Answer:

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Explanation:

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For the second case, we consider the characteristics to now be

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Recall that the kinetic energy of a spinning body is given as

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and for the second case, the K.E. is given as

KE = \frac{1}{2}(5I)(\frac{w}{5} )^{2}   = \frac{5}{50}Iw^{2}

KE = \frac{1}{10}Iw^{2}

<em>this is one-tenth the kinetic energy before its spinning characteristics were changed.</em>

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3 years ago
A 90kg person jumps from a 30m tower into a tub of water with a volume of 5m3 initially at 20°C. Assuming that all of the work d
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Eₐ = 1000 kg / m³ * 5 m³ * 9.8 m / s²

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