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4 years ago

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a child is riding a bike at a speed of 0.6m/s with a total kinetic energy of 12.4J.If the mass of the child is 30kg, what is the

mass of the bike?

2 answers:

Nesterboy [21]4 years ago

8 0

Ek=kinetic energy (J)
m=mass of object (Kg)
v=speed of object (m/s)

We have the following formula:

Ek=(1/2)mv² ⇒ m=2(Ek)/v²

In this case:
m= mass of the child + mass of the bike
m=30 kg + m₁ (m₁=mass of the bike).

We would have to modify our formula like this:

m=2(Ek)/v²
m=30 kg + m₁

30 kg +m₁=2(Ek)/v²
m₁=2(Ek)/v²-30 kg

We have the following data:

v=0.6 m/s
Ek=12.4J
m=30 Kg + m₁

Therefore:

m₁=2(Ek)/v² - 30 kg
m₁=2(12.4 J)/(0.6 m/s)² - 30 kg
m₁=24.8 J/(0.36 m²/s²) -30 kg
m₁=68.89 Kg - 30 kg
m₁=38.89 Kg

Answer: the mass of the bike is 38.89 Kg.

vova2212 [387]4 years ago

8 0

The mass of the bike is 38.89 Kg.

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3 years ago

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This situation is related to projectile motion or parabolic motion, in which the travel of the cannonball has two components: x-component and y-component. Being their main equations as follows:

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Where:

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$\theta=0$ because we are told the cannonball is shot horizontally

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Where:

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$y=0$ is the final height of the cannonball (when it finally hits the ground)

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We need to find how far (horizontally) the cannonball has traveled before landing. This means we need to find the maximum distance in the x-component, let's call it $X_{max}$ and this occurs when $y=0$ .

So, firstly we will find the time with (2):

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Rearranging the equation:

$0=-(4.9 m/s^{2})t^2+48. 1 m/s sin(0\°) t+1.5 m$ (4)

$-(4.9 m/s^{2})t^2+(48. 1 m/s) t+1.5 m=0$ (5)

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Where:

$a=-4.9 m/s^{2}$

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Solving (7) we find the positive result is:

$t=9.847 s$ (8)

Substituting this value in (1):

$x=(48.1 m/s)cos(0\°) (9.847 s)$ (9)

$x=473.640 m$ This is the horizontal distance the cannonball traveled before it landed on the ground.

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