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3 years ago

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3. An object with a mass of 10 kg is moving along a horizontal surface. At a certain point it has 40 J of kinetic energy. If the

coefficient of friction between the object and the surface is 0.40, how far will the object go beyond that point before coming to a stop? Use g = 10 m/s2

1 answer:

damaskus [11]3 years ago

7 0

Answer: 100cm

Explanation:

The force of friction on a surface normal to gravity where µ is the coefficient of friction is

F = µmg

Where

F = the friction force

µ = coefficient of friction

m = mass of the object

g = acceleration due to gravity

Also, the Kinetic Energy of the object, E = Fs, where

E = Kinetic Energy

s = stopping distance. So that,

E = µmgs

40 J = 0.4 * 10 kg * 10 m/s² * s

40 J = 40 kgm/s² * s

s = 40 J / 40 kgm/s²

s = 1 m or 100 cm

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Where $T_{1}$ & $T_{2}$ represents the compressor inlet and outlet temperature.

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This is the value of temperature at compressor outlet.

The temperature ratio inside the turbine is given by

$\frac{T_{3} }{T_{4} }$ = $r_{p}\frac{\gamma - 1}{\gamma}$

Where $T_{3}$ & $T_{4}$ represents the turbine inlet & outlet temperatures.

Put all the values in the given formula we get

⇒ $\frac{1400}{T_{4} }$ = $8^{\frac{1.4 - 1}{1.4} }$

⇒ $T_{4}$ = 773.05 K

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(b). The work done by the turbine is given by the formula ( $W_{T}$ ) = $C_{p}$ ( $T_{3}$ - $T_{4}$ )

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And work done inside the pump is given by $W_{c}$ = $C_{p}$ ( $T_{2}$ - $T_{1}$ )

Put all the values in the above formula we get $W_{c}$ = 1.005 × (543.43 - 300)

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Back work ratio is given by $r_{b}$ = $\frac{W_{C} }{W_{}T}$

Put the values of $W_{c}$ & $W_{T}$ in above formula we get

⇒ $r_{b}$ = $\frac{244.65}{630.08}$

⇒ $r_{b}$ = 0.388

This is the value of back work ratio.

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⇒ E = 1 - $\frac{1}{8 ^{\frac{1.4 - 1}{1.4} } }$

⇒ E = 0.448

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