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aleksley [76]
3 years ago
9

Think of a skateboarder riding down a hill. Once the skateboarder reaches the bottom of the hill, if she does not manually add m

otion, the skateboarder will slow until she stops. Why does the skateboarder eventually stop?
Physics
2 answers:
ivanzaharov [21]3 years ago
8 0

From first law of motion, law of inertia, we know that an object remains in the state of rest or motion unless acted upon by an unbalanced external force. The skateboard riding down a hill when reaches the bottom slows down due to <u>frictional force </u>acting in the opposite direction. Frictional force is a contact force which acts between two surfaces against the direction of  motion of one surface over the other. Here, frictional force acts <u>between skateboard and ground</u>. This reduces the speed of the skateboard and it eventually stops.

LUCKY_DIMON [66]3 years ago
7 0
The wheels give off energy to the ground because of the friction until they run out, think of the wheels as a battery and the ground as a device that needs power, it just sucks out the power
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Answer:

The box stops in 0.139 seconds, after moving 7.29cm (0.0729m) backwards relative to the belt.

Explanation:

As the box is initially at rest relative to the earth, it is moving backwards with a speed of 1.05m/s relative to the belt. Then, the frictional force acts on the box to make it stop relative to the belt. So, we first have to write the equations of motion of the box in each axis:

x: f_k=ma\implies a=\frac{f_k}{m} \\\\y: N-mg=0\implies N=mg

Since the frictional force f_k is equal to f_k=\mu_k N=\mu_k mg, then we have that the acceleration is:

a=\frac{\mu_k mg}{m}=\mu_k g

Now, from the definition of acceleration we get:

a=\frac{v-v_0}{t}\implies t=\frac{v-v_0}{a}

And, as the final velocity is zero because the box gets to a stop, we have:

t=-\frac{v_0}{a}=-\frac{v_0}{\mu_k g}

(Don't worry about the negative sign. It will disappear because the initial velocity is also negative, since we take the box initially moving backwards)

Then, plugging in the given values, we calculate the time:

t=-\frac{(-1.05m/s)}{0.770(9.81m/s^{2})}=0.139s

In words, the time the box takes to stop sliding relative to the belt is 0.139s.

The displacement of the box in this time, is given by the kinematics formula:

v^{2}=v_0^{2}+2ax\implies x=-\frac{v_0^{2}}{2\mu_kg}

Finally, we calculate the displacement:

x=-\frac{(1.05m/s)^{2} }{2(0.770)(9.81m/s^2)}=-0.0729m=-7.29cm

This means that the box moves 7.29cm backwards relative to the belt.

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where, all alphabets are in their usual meanings.

Now, we shall calculate the change in potential energy of the stone

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or,                = 0.5 kg ×9.8 m/s² ×(2.0 m - 1.0 m)

or,                = 4.9 J

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