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Serga [27]
3 years ago
9

With what minimum speed must you toss a 200 g ball straight up to just touch the 14-m-high roof of the gymnasium if you release

the ball 1.3 m above the ground? solve this problem using energy.
Physics
1 answer:
Brut [27]3 years ago
8 0
When they say use energy, you want to use

Total energy = potential energy + kinetic energy = mgh + 1/2mv²

I assume you mean 200 g ball,

so, we know the total distance traveled is going to be 13 - 1.3 = 11.7 m

If the ball just makes it to the top ( 13 m ) , then the ball will stop moving and the kinetic energy will be 0,

therefore, the potential energy at the top will be the total energy of the system = mgh

from this, we say that mgh = 1/2mv² solve for v

<span> v = sqrt (2gh) = 15.2 m/s </span>
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Answer:

21.8°

Explanation:

Let's call θ the angle between BC and the horizontal.

Draw a free body diagram for each block.

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and tension force T pushing parallel up AB.

There are 4 forces acting on block E:

Weight force P pulling down,

Normal force N₂ pushing perpendicular to BC,

Friction force N₂μ pushing parallel to BC,

and tension force T pulling parallel to BC.

Sum of forces on D in the perpendicular direction:

∑F = ma

N₁ − P sin θ = 0

N₁ = P sin θ

Sum of forces on D in the parallel direction:

∑F = ma

T + N₁μ − P cos θ = 0

T = P cos θ − N₁μ

T = P cos θ − P sin θ μ

T = P (cos θ − sin θ μ)

Sum of forces on E in the perpendicular direction:

∑F = ma

N₂ − P cos θ = 0

N₂ = P cos θ

Sum of forces on E in the parallel direction:

∑F = ma

N₂μ + P sin θ − T = 0

T = N₂μ + P sin θ

T = P cos θ μ + P sin θ

T = P (cos θ μ + sin θ)

Set equal:

P (cos θ − sin θ μ) = P (cos θ μ + sin θ)

cos θ − sin θ μ = cos θ μ + sin θ

1 − tan θ μ = μ + tan θ

1 − μ = tan θ μ + tan θ

1 − μ = tan θ (μ + 1)

tan θ = (1 − μ) / (1 + μ)

Plug in values:

tan θ = (1 − 0.4) / (1 + 0.4)

θ = 23.2°

∠BCA = 45°, so the angle of AC relative to the horizontal is 45° − 23.2° = 21.8°.

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Answer:

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