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Vilka [71]
3 years ago
6

An archer puts a 0.30 kg arrow to the bowstring. An average force of 201 N is exerted to draw the string back 1.3 m.a. Assuming

that all the energy goes into the arrow, with what speed does the arrow leave the bow? (41.7 m/s)b. If the arrow is shot straight up, how high does it rise? (88.9 m)
Physics
1 answer:
Vlad [161]3 years ago
4 0

Answer:

Explanation:

Given

mass of archer m=0.3\ kg

Average force F_{avg}=201\ N

extension in arrow x=1.3\ m

Work done to stretch the bow with arrow

W=F\cdot x

W=201\times 1.3=261.3\ m

This work done is converted into kinetic Energy of arrow

W=\frac{1}{2}mv^2

where v= velocity of arrow

261.3=\frac{1}{2}\times 0.3\times v^2

v=\sqrt{1742}

v=41.73\ m/s

(b)if arrow is thrown vertically upward then this energy is converted to Potential energy

W=mgh

261.3=0.3\times 9.8\times h

h=\frac{261.3}{0.3\times 9.8}

h=88.87\ m

   

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Answer:

a. 79.1 N

b. 344 J

c. 344 J

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Explanation:

a. Since the crate has a constant velocity, its net force must be 0 according to Newton's 1st law. The push force F_p by the worker must be equal to the friction force F_f on the crate, which is the product of friction coefficient μ and normal force N:

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F_p = F_f = \mu N = \mu mg = 0.26 * 31 * 9.81 = 79.1 N

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c. The work is done on the crate by friction force is also the product of friction force and the distance traveled s = 4.35

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e. The total work done on the crate would be sum of the work done by the pushing force and the work done by friction

W_p + W_f = 344 - 344 = 0 J

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