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3 years ago

44. ( John weighs 80 kg, and eats a 1,000 J candy bar. If his

Physics

1 answer:

Darina [25.2K]3 years ago

4 0

John can run with the velocity of 5 m/s

Explanation:

Kinetic energy is defined as the energy is being used to do an activity, basically energy associated with the motion of objects in the universe.
The formula used to find the kinetic energy of an object is k = $\frac{1}{2} mv^{2}$ where as k represented as kinetic energy, m is the mass of the object and v is the velocity of the given object.

Here, to find the answer we have to re-write the equation as $v = \sqrt[2]{\frac{2 k}{m} }$
Given, the mass of the object, here it is John = 80 kg, energy needs to be converted to kinetic energy, k = 1000 J.
Hence, substitute all the values, then you would velocity as 5 m/s

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A 2.0 kg wood block is launched up a wooden ramp that is inclined at a 30˚ angle. The block’s initial speed is 10 m/s. What vert

WARRIOR [948]

Answer:

The wood block reaches a height of 4.249 meters above its starting point.

Explanation:

The block represents a non-conservative system, since friction between wood block and the ramp is dissipating energy. The final height that block can reach is determined by Principle of Energy Conservation and Work-Energy Theorem. Let suppose that initial height has a value of zero and please notice that maximum height reached by the block is when its speed is zero.

$\frac{1}{2}\cdot m \cdot v^{2} = m \cdot g\cdot h + \mu_{k}\cdot m\cdot g\cdot s \cdot \sin \theta$

$\frac{1}{2}\cdot v^{2} = g\cdot h + \mu_{k}\cdot g\cdot \left(\frac{h}{\sin \theta} \right)\cdot \sin \theta$

$\frac{1}{2}\cdot v^{2} = g\cdot h +\mu_{k}\cdot g\cdot h$

$\frac{1}{2}\cdot v^{2} = (1 +\mu_{k})\cdot g\cdot h$

$h = \frac{v^{2}}{2\cdot (1 + \mu_{k})\cdot g}$ (1)

Where:

$h$ - Maximum height of the wood block, in meters.

$v$ - Initial speed of the block, in meters per second.

$\mu_{k}$ - Kinetic coefficient of friction, no unit.

$g$ - Gravitational acceleration, in meters per square second.

$m$ - Mass, in kilograms.

$s$ - Distance travelled by the wood block along the wooden ramp, in meters.

$\theta$ - Inclination of the wooden ramp, in sexagesimal degrees.

If we know that $v = 10\,\frac{m}{s}$ , $\mu_{k} = 0.20$ and $g = 9.807\,\frac{m}{s^{2}}$ , then the height reached by the block above its starting point is:

$h = \frac{\left(10\,\frac{m}{s} \right)^{2}}{2\cdot (1+0.20)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}$

$h = 4.249\,m$

The wood block reaches a height of 4.249 meters above its starting point.

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