Respuesta :
Answer:
Speed of the arrow, v = 43.04 m/s
Explanation:
It is given that,
Mass of an arrow, m = 85 g = 0.085 kg
Average force exerted by the string of a bow, F = 105 N
Distance covered, d = 75 cm = 0.75 m
Initial speed of then arrow, u = 0
To find,
The speed of the arrow as it leaves the bow.
Solution,
Let u and v is the initial and the final speed of the arrow. We need to find the the speed of the arrow as it leaves the bow. Using work energy theorem to find it as :
Work done = Final kinetic energy - initial kinetic energy
[tex]F.d=\dfrac{1}{2}mv^2[/tex]
[tex]v=\sqrt{\dfrac{2Fd}{m}}[/tex]
[tex]v=\sqrt{\dfrac{2\times 105\times 0.75}{0.085}}[/tex]
v = 43.04 m/s
Therefore, the speed of the arrow as it leaves the bow is 43.04 m/s.
The speed of the arrow as it leaves the bow is 43.04 m/sec.
Given :
Mass of an arrow, m = 85 g = 0.085 Kg
Average force, F = 105 N
Distance covered, s = 75 cm = 0.75 m
Initial speed of arrow, u = 0
Solution :
We know that,
Work Done = Final Kinetic Energy - Initial Kinetic Energy
here, Initial Kinetic Energy = 0 because initial velocity is zero. Therefore
[tex]\rm W = \dfrac{1}{2}mv^2[/tex] ---- (1)
where, v is final velocity, m is mass and W is work done.
We also know that,
Work Done = Force x Displacement
[tex]\rm W = F\times s[/tex] ---- (2)
From equation (1) and (2) we get,
[tex]\rm F\times s = \dfrac{1}{2}mv^2[/tex]
[tex]\rm v = \sqrt{\dfrac{2\times F \times s}{m}}[/tex] ----- (3)
Now put the values of F, s and m in equation (3) we get,
[tex]\rm v= \sqrt{\dfrac{2\times 105 \times 0.75}{0.085}}[/tex]
[tex]\rm v = 43.04\;m/sec[/tex]
Therefore, the speed of the arrow as it leaves the bow is 43.04 m/sec.
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https://brainly.com/question/17127206?referrer=searchResults
