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andrew-mc [135]

2 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]2 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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8. A 3.0 x 10' g nerf projectile is fired from a 1.5 kg nerf gun. If the velocity of the projectile is

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Answer:

$v_g=0.2\ m.s^{-1}$

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Answer:

<em>The third piece moves at 6.36 m/s at an angle of 65° below the horizon</em>

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Linear Momentum

It's a physical magnitude that measures the product of the velocity by the mass of a moving object. In a system where no external forces are acting, the total momentum remains unchanged regardless of the interactions between the objects in the system.

If the velocity of an object of mass m is $\vec v$ , the linear momentum is computed by

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Since the board was initially at rest

$\displaystyle \vec{P}_{t1}=$

After the explosion, 3 pieces are propelled in different directions and velocities, and the total momentum is

$\displaystyle \vec{P}_{t2}=m_1\ \vec{v}_1\ +\ m_2\ \vec{v}_2+m_3\ \vec{v}_3$

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$\displaystyle \vec{v}_1=(10\ m/s,60^o)$

We find the components of that velocity

$\displaystyle \vec{v}_1=$

$\displaystyle \vec{v}_1=m/s$

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$\displaystyle \vec{v}_2=(15,180^o)$

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$\displaystyle \vec{v}_2=(-15,0)\ m/s$

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$\displaystyle P_{t2}=2+1.2+m_3\ \vec{v}_3$

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Knowing the total momentum equals the initial momentum

$\displaystyle P_{t2}=+m_3\ \vec{v}_3=0$

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$\displaystyle m_3\ \vec{v}_3=$

Calculating

$\displaystyle m_3\ \vec{v}_3=$

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b)

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$\displaystyle \vec{v}_3=\frac{1}{3}$

$\displaystyle \vec{v}_3=m/s$

The magnitude of the velocity is

$\displaystyle \vec{v}_3|=\sqrt{2.67^2+(-5.77)^2}=6.36$

And the angle is

$\displaystyle tan\theta =\frac{-5.77}{2.67}=-2.161$

$\displaystyle \theta =-65.17^o$

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