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zmey [24]
3 years ago
10

Starting from rest, a 2.1x10-4 kg flea springs straight upward. While the flea is pushing off from the ground, the ground exerts

an average upward force of 0.41 N on it. This force does 3.7x10-4 J of work on the flea. (a) What is the flea's speed when it leaves the ground? (b) How far upward does the flea move while it is pushing off? Ignore both air resistance and the flea's weight.
Physics
1 answer:
nirvana33 [79]3 years ago
8 0

Answer:

1.327363 m/s

0.00090243 m

Explanation:

u = Initial velocity

v = Final velocity

m = Mass of flea

Energy

E=\frac{1}{2}m(v^2-u^2)\\\Rightarrow 3.7\times 10^{-4}=2.1\times 10^{-4}(v^2-0)\\\Rightarrow v=\sqrt{\frac{3.7\times 10^{-4}}{2.1\times 10^{-4}}}\\\Rightarrow v=1.32736\ m/s

The velocity of the flea when leaving the ground is 1.327363 m/s

W=F\times s\\\Rightarrow s=\frac{W}{F}\\\Rightarrow s=\frac{3.7\times 10^{-4}}{0.41}\\\Rightarrow s=0.00090243\ m

The flea will travel 0.00090243 m upward

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A mass m = 1.1 kg hangs at the end of a vertical spring whose top end is fixed to the ceiling. The spring has spring constant k
adell [148]

Answer:

a) = 10.22 rad/s

b) = 0.35 m

Explanation:

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6 0
3 years ago
A 72.9-kg base runner begins his slide into second base when moving at a speed of 4.02 m/s. The coefficient of friction between
elena-14-01-66 [18.8K]

Answer:

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Explanation:

Using work-kinetic energy theorem, the work done by friction = kinetic energy change of the base runner

So, W = ΔK

W = 1/2m(v₁² - v₀²) where m = mass of base runner = 72.9 kg, v₀ = initial speed of base runner = 4.02 m/s and v₁ = final speed of base runner = 0 m/s(since he stops as he reaches home base)

So, substituting the values of the variables into the equation, we have

W = 1/2m(v₁² - v₀²)

W = 1/2 × 72.9 kg((0 m/s)² - (4.02 m/s)²)

W = 1/2 × 72.9 kg(0 m²/s² - 16.1604 m²/s²)

W = 1/2 × 72.9 kg(-16.1604 m²/s²)

W = 1/2 × (-1178.09316 kgm²/s²)

W = -589.04658 kgm²/s²

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