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True [87]
3 years ago
5

An archer puts a 0.285 kg arrow in a bow and uses an average force of 182 N to draw the string back 1.32 m. Assume the energy st

ored in the bow is transferred to the arrow when it is shot. (a) What is the speed of the arrow as it leaves the bow
Physics
1 answer:
vagabundo [1.1K]3 years ago
4 0

Answer:

speed of the arrow as it leaves is 41.05 m/sec

Explanation:

We have given mass m = 0.285 kg

Average force F = 182 N

Distance traveled d = 1.32 m

We know that work done = force \timesdistance

Sp work done = 182\times 1.32=240.24J

Now according to  work energy theorem work done will be equal to kinetic energy

So \frac{1}{2}mv^2=240.24

\frac{1}{2}\times 0.285\times v^2=240.24

v^2=1685.89

v = 41.05 m/sec

So speed of the arrow as it leaves is 41.05 m/sec

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The driver of a pickup truck accelerates from rest to a speed of 37 mi/hr over a horizontal distance of 215 ft with constant acc
ZanzabumX [31]

Answer:

Maximum shearing force developed in each of the two pegs during acceleration is 1830 lbf

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First convert speed into ft/sec

1 mile/hr = 1.47 ft/sec

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37 mile/hr = 37 x 1.47 ft/sec

37 mile/hr =  54.39 ft/sec

with initial speed 0 ft/sec (starting from rest), using in equation of motion:

a = [(54.39 ft/sec)² - (0 ft/sec)²]/2(215 ft)

a = 6.88 ft/sec²

Now, the total shear force will be given by Newton's second law of motion:

F = ma

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F = 3660 lbf

Now for the max shear force in each of the two pegs we divide total fore by 2:

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Answer:

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