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Hoochie [10]
3 years ago
11

Mario, a hockey player, is skating due south at a speed of 4.79 m/s relative to the ice. A teammate passes the puck to him. The

puck has a speed of 9.13 m/s and is moving in a direction of 17.8 ° west of south, relative to the ice. What are (a) the magnitude and (b) direction (relative to due south) of the puck's velocity, as observed by Mario
Physics
1 answer:
frez [133]3 years ago
5 0

Mario's velocity vector relative to the ice is

\vec v_{M/I}=-4.79\,\vec\jmath

(note that all velocities mentioned here are given in m/s)

The puck is moving in a direction of 17.8 degrees west of south, or 252.2 degrees counterclockwise relative to east. Its velocity vector relative to the ice is then

\vec v_{P/I}=9.13(\cos252.2^\circ\,\vec\imath+\sin252.2^\circ\,\vec\jmath)=-2.79\,\vec\imath-8.69\,\vec\jmath

The velocity of the puch relative to Mario is

\vec v_{P/M}=\vec v_{P/I}+\vec v_{I/M}=\vec v_{P/I}-\vec v_{M/I}

\vec v_{P/M}=-2.79\,\vec\imath-3.90\,\vec\jmath

Then, relative to Mario,

a. the puck is traveling at a speed of \boxed{\|\vec v_{P/M}\|=4.80}, and

b. is moving in a direction \theta such that

\tan\theta=\dfrac{-3.90}{4.80}\implies\theta=-126^\circ

which is about \boxed{35.6^\circ} west of south.

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Answer:

6 N

Explanation:

Let's start with the small block m on top.  There are four forces:

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Now let's look at the large block M on bottom.  There are seven forces:

Normal force N₁ pushing down (opposite and equal from block m),

Friction force N₁μ pushing right (opposite and equal from block m),

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Tension force T pulling right,

Applied force F pulling left,

Normal force N₂ pushing up,

and friction force N₂μ pushing right (opposing the direction of motion).

So you've correctly identified the free body diagrams.

Now apply Newton's second law.  Sum of forces in the y direction for block m:

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N₁ = mg

Sum of forces in the x direction:

∑F = ma

T − N₁μ = 0

T = N₁μ

T = mgμ

Sum of forces in the y direction for block M:

∑F = ma

-N₁ − Mg + N₂ = 0

N₂ = N₁ + Mg

N₂ = mg + Mg

Sum of forces in the x direction:

∑F = ma

N₁μ + T − F + N₂μ = 0

F = N₁μ + T + N₂μ

F = mgμ + mgμ + (mg + Mg)μ

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F = 5gμm

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A) x=\pm \frac{A}{2\sqrt{2}}

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The total energy, which is conserved, at any other point of the motion is the sum of elastic potential energy and kinetic energy:

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We want to know the displacement x at which the elastic potential energy is 1/3 of the kinetic energy:

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