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Helga [31]
4 years ago
14

A wood block, after being given a starting push, slides down a ramp at a constant speed. what is the angle of the ramp above hor

izontal? suppose that μk = 0.35.
Physics
1 answer:
Stells [14]4 years ago
5 0

The solution for the problem is:

Constant speed means Fnet = 0. 
Let m = mass of wood block and Θ = angle of ramp; then if µk = 0.35 …

The computation would be:


Fnet = 0 = mg (sin Θ) - (µk) (mg) (cos Θ) 
mg (sin Θ) = µk (mg) (cos Θ) 
µk = tan Θ 
Θ = arctan(µk)

= arctan (0.35)

≈ 19.3°

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I assume that the ball is stationary (v=0) at point B, so its total energy is just potential energy, and it is equal to 7.35 J. 
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And since K=7.35 J, we can find the velocity, v:
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A quarterback passes a football from height h = 2.1 m above the field, with initial velocity v0 = 13.5 m/s at an angle θ = 32° a
SOVA2 [1]

Answer:

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This is a projectile launching exercise, let's use trigonometry to find the components of the initial velocity

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        cos θ = v₀ₓ / vo

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the time the ball is in the air is twice the time to reach the maximum height, where the vertical speed is zero

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       x = v₀ cos θ (2 v_{o} sin θ / g)

       x = v₀² /g      2 cos θ sin θ

       x = v₀² sin 2θ / g

at the point where the receiver receives the ball is at the same height, so this coincides with the range of the projectile launch,

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at the highest point the vertical speed is zero

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           t = v_{oy} / g

           t = v₀ sin θ / g

as the time to get on and off is the same the total time or flight time is

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           t_total = 2 v₀ sin θ / g

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          x = 16.7 m

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