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andrew-mc [135]

3 years ago

6

A 15.0 kg crate, initially at rest, slides down a ramp 2.0 m long and inclined at an angle of 20.0° with the horizontal. Using t

he work-kinetic energy theorem and disregarding friction, find the
velocity of the crate at the bottom of the ramp. (g = 9.81 m/s?)

1 answer:

Elis [28]3 years ago

3 0

The component of the crate's weight that is parallel to the ramp is the only force that acts in the direction of the crate's displacement. This component has a magnitude of

<em>F</em> = <em>mg</em> sin(20.0°) = (15.0 kg) (9.81 m/s^2) sin(20.0°) ≈ 50.3 N

Then the work done by this force on the crate as it slides down the ramp is

<em>W</em> = <em>F d</em> = (50.3 N) (2.0 m) ≈ 101 J

The work-energy theorem says that the total work done on the crate is equal to the change in its kinetic energy. Since it starts at rest, its initial kinetic energy is 0, so

<em>W</em> = <em>K</em> = 1/2 <em>mv</em> ^2

Solve for <em>v</em> :

<em>v</em> = √(2<em>W</em>/<em>m</em>) = √(2 (101 J) / (2.0 m)) ≈ 10.0 m/s

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Answer:

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Given:

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We have to find the value of $R$ we know that in series circuit current are same.

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4 0

3 years ago

A 108 kg clock initially at rest on a horizontal floor requires a 653 N horizontal force to set it in motion. After the clock is

kifflom [539]

Answer:

$\mu_s=0.61$

$\mu_k=0.49$

Explanation:

Given that,

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For the clock in motion, force acting on it, $F_2=527\ N$

To find,

$\mu_s\ and\ \mu_k$

Solution,

When an object is at rest, the force acting on it is called force due to static friction and if the object is in motion, the force acting on its called force due to kinetic friction.

Let $\mu_s$ is the coefficient of static friction. Force is given by :

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Let $\mu_k$ is the coefficient of kinetic friction. Force is given by :

$F_k=\mu_kmg$

$\mu_k=\dfrac{F_k}{mg}$

$\mu_k=\dfrac{527}{108\times 9.8}$

$\mu_k=0.49$

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