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Goshia [24]
3 years ago
8

A 87 arrow is fired from a bow whose string exerts an average force of 105 on the arrow over a distance of 75 .

Physics
1 answer:
timofeeve [1]3 years ago
6 0

The solution would be like this for this specific problem:

 

V^2 = 2AS = 2FS/M

V = sqrt(2FS/M) = sqrt(2*105*.75/.087) = 44.52817783 = 42.5 mps

So the speed of the arrow as it leaves the bow is 42.5 mps.

I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

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When an average force F is exerted over a certain distance on a shopping cart of mass m, its kinetic energy increases by 12mv2.
VMariaS [17]

Answer:

A)d=\dfrac{1}{2F}mv^2

B)\Delta KE'=2\times \dfrac{1}{2}mv^2

Explanation:

Given that

Force  = F

Increase in Kinetic energy = \dfrac{1}{2}mv^2

\Delta KE=\dfrac{1}{2}mv^2

we know that

Work done by all the forces =change in the kinetic energy

a)

Lets distance = d

We know work done by force F

W= F .d

F.d=ΔKE

F.d=\dfrac{1}{2}mv^2

d=\dfrac{1}{2F}mv^2

b)

If the force become twice

F' = 2 F

F'.d=ΔKE'

2 F .d = ΔKE'                          ( F.d =Δ KE)

2ΔKE = ΔKE'

\Delta KE'=2\times \dfrac{1}{2}mv^2

Therefore the final kinetic energy will become the twice if the force become twice.

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Read 2 more answers
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