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Dvinal [7]
3 years ago
14

A marble is dropped from rest and falls for 2.3 seconds. Find its final velocity.

Physics
1 answer:
juin [17]3 years ago
4 0

Answer:

23 m/s downward

__________________________________________________________

<em>Taking the downward direction as positive</em>

<u>We are given:</u>

Initial velocity of the marble (u) = 0 m/s

Time interval (t) = 2.3 seconds

Final velocity (v) = x m/s

<u>Solving for the Final velocity:</u>

<u>Acceleration of the Marble:</u>

We know that gravity will make the marble accelerate at a constant acceleration of 10 m/s

<u>Final velocity:</u>

v = u + at                                              [First equation of motion]

x = 0 + (10)(2.3)                                    [replacing the given values]

x = 23 m/s

Hence, after 2.3 seconds, the marble will move at a velocity of 23 m/s in the downward direction

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When energy is added to an object, it will vibrate at it's...
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he block is released, and it slides 2.0 m (from the point at which it is released) across a horizontal surface before friction s
alex41 [277]

Answer:

0.245

Explanation:

When the block is released, the initial elastic potential energy stored in the spring is entirely converted into kinetic energy of the block.

Therefore, we can calculate the initial speed of the block:

\frac{1}{2}kx^2 = \frac{1}{2}mv^2

where the term on the left is the potential energy and where the term on the right is the kinetic energy, and where

k = 4500 N/m is the spring constant

x = 8.0 cm = 0.08 m is the compression of the spring

m = 3.0 kg is the mass of the block

v is the initial velocity

Solving for v,

v=\sqrt{\frac{kx^2}{m}}=\sqrt{\frac{(4500)(0.08)^2}{3.0}}=3.1 m/s

Then, after the block is released, all its kinetic energy is converted into thermal energy as the block slows down, due to friction. Therefore, the work done by friction is equal to the initial kinetic energy of the block.

The force of friction is

F=\mu mg

where

\mu is the coefficient of friction

g=9.8 m/s^2 is the acceleration of gravity

So the work done by it is (in magnitude)

W=Fd=\mu mg d

where

d = 2.0 m is the distance covered

Therefore,

\frac{1}{2}mv^2 = \mu mg d

And solving for \mu,

\mu = \frac{v^2}{2gd}=\frac{3.1^2}{2(9.8)(2.0)}=0.245

8 0
3 years ago
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