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Liula [17]
2 years ago
11

A swift blow with the hand can break a pine board. As the hand hits the board, the kinetic energy of the hand is transformed int

o elastic potential energy of the bending board; if the board bends far enough, it breaks. Applying a force to the center of a particular pine board deflects the center of the board by a distance that increases in proportion to the force. Ultimately the board breaks at an applied force of 870 N and a deflection of 1.4 cm.
a. To break the board with a blow from the hand, how fast must the hand be moving? Use 0.50 kg for the mass of the hand.
b. If the hand is moving this fast and comes to rest in a distance of 1.2 cm, what is the average force on the hand?
Physics
1 answer:
Stells [14]2 years ago
3 0

Answer:

A. The hand must move with a velocity of 6.98 m/s to break the board.

B. Average force on the hand = 1025 N

Explanation:

A.To determine the speed the hand must move with to break the board, the force workdone in breaking the board is found first.

Workdone = force × distance

Minimum force required = 870 N;

Distance moved by board/Deflection in order to break = 1.4 cm = 0.014 M

Workd done = 870 N × 0.014 m = 12.18 Nm or 12.18 J

This work done = Kinetic energy of the hand

Kinetic energy = mv²/2 ; where m is mass and v is velocity

Mass of hand = 0.50 Kg, velocity = ?, K.E. = 12.18 J

v² = 2 KE/m

v = √2KE/m

v = √(2 × 12.18/0.50)

v = 6.98 m/s

Therefore, the hand must move with a velocity of 6.98 m/s to break the board.

B. Average force on the hand

This can be determined using the equation of motion, v² = u² + 2as to find acceleration, since force = mass × acceleration

From the equation of motion, a = v² - u²/2s

At rest, v = 0, u = 6.98, s = 1.2 cm = 0.012 m

a = 0² - 6.98²/ 2 × 0.012

a = -2030 m/s²

Force = 2030 m/s² × 0. 50 kg = 1015 N

Therefore, Average force on the hand = 1025 N

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Answer:

a)v=1.77\times 10^6\ m/s

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Explanation:

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m a =  q .V/d b            

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Now by putting the all values in equation 1

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v^2=u^2+2as

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s = 0.1 cm

v^2=(4\times 10^5)^2+2\times 1.5\times 10^{15}\times 0.1\times 10^{-2}

v=\sqrt{3.16\times 10^{12}}\ m/s

v=1.77\times 10^6\ m/s

b)

s = 0.5 cm

v^2=(4\times 10^5)^2+2\times 1.5\times 10^{15}\times 0.5\times 10^{-2}

v=\sqrt{1.5\times 10^{13}}\ m/s

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s = 1 cm

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s = 1.5 cm

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e)

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v=\sqrt{6.06\times 10^{13}}\ m/s

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