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NeTakaya
3 years ago
9

An arrow, starting from rest, leaves the bow with a speed of25

Physics
1 answer:
xz_007 [3.2K]3 years ago
5 0

Answer:

  • The speed will be v_f= 43.30 \frac{m}{s}

Explanation:

We can use the following kinematics equation

v_f^2-v_i^2=2 \ a \ d

where v_f is the final speed, v_i its the initial speed, a is the acceleration, and d the distance.

The force will be tripled, the force is:

\vec{F} = m \vec{a}

in 1D

F = m a

Now, for the original problem, we have

(25 \frac{m}{s})^2-(0 \frac{m}{s})^2=2 \ a' \ d'

(25 \frac{m}{s}))^2=2 \ a' \ d'

625 \frac{m^2}{s^2}=2 \ a' \ d'

For the second problem, we have

(v_f})^2-(v_i)^2=2 \ a'' \ d''

starting from the rest, we have the same initial velocity.

(v_f})^2-(0\frac{m}{s})^2=2 \ a'' \ d''

(v_f})^2=2 \ a'' \ d''

As the force is tripled, we have:

F'' = 3 F'

m'' \ a'' = 3 m' \ a'

But the mass its the same,  so

m' \ a'' = 3 m' \ a'

a'' = 3 \ a'

So the acceleration its also tripled.

(v_f})^2=2 \ (3 * a') \ d''

(v_f})^2=3 ( 2 \ ( a' \ d'' )

As the distance traveled by the arrow must also be the same, we have:

(v_f})^2=3 ( 2 \ ( a' \ d' )

(v_f})^2=3 (625 \frac{m^2}{s^2})

v_f= \ sqrt{3 (625 \frac{m^2}{s^2})}

v_f= \ sqrt{3} 25 \frac{m}{s}

v_f= 43.30 \frac{m}{s}

And this will be the speed from the arrow leaving the bow.

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