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melomori [17]
3 years ago
7

A block of ice of mass 4.30 kg is placed against a horizontal spring that has a force constant k = 250 N/m and is compressed a d

istance 2.80 × 10⁻² m. The spring is released and accelerates the block along a horizontal surface. You can ignore friction and the mass of the spring.
Calculate the work done on the block by the spring during the motion of the block from its initial position to where the spring has returned to its uncompressed length. What is the speed of the block after it leaves the spring?
Physics
1 answer:
OleMash [197]3 years ago
8 0

Answer:

W = 0.060 J

v_2 = 0.18 m/s

Explanation:

solution:

for the spring:

W = 1/2*k*x_1^2 - 1/2*k*x_2^2

x_1 = -0.025 m and x_2 = 0

W = 1/2*k*x_1^2 = 1/2*(250 N/m)(-0.028m)^2

W = 0.060 J

the work-energy theorem,

W_tot = K_2 - K_1 = ΔK

with K = 1/2*m*v^2

v_2 = √2*W/m

v_2 = 0.18 m/s

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If sea otters are hunted and their population is brought to extinction then this will cause major harm the ecosystem of the kelp forest and it will disturb the ecological balance in the kelp forest. The herbivorous animals consumed by the sea otters will increase in population and they will consume a lot of vegetation in the forest. The kelp forest which forms the coastline will not remain effective in providing protection against the storms to the neighboring areas.

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5595 m

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2 years ago
A spherical shell of radius 3.59 cm and a cylinder of radius 7.22 cm are rolling without slipping along the same floor. The two
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Answer:

(ω₁ / ω₂) = 1.9079

Explanation:

Given

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We know that

K₁ = Kt₁ + Kr₁ = 0.5*m₁*v₁²+0.5*I₁*ω₁²

if

v₁ = ω₁*R₁

and

I₁ = (2/3)*m₁*R₁² = (2/3)*m*R₁²

∴    K₁ = 0.5*m*ω₁²*R₁²+0.5*(2/3)*m*R₁²*ω₁²   <em>(I)</em>

then

K₂ = Kt₂ + Kr₂ = 0.5*m₂*v₂²+0.5*I₂*ω₂²

if

v₂ = ω₂*R₂

and

I₂ = 0.5*m₂*R₂² = 0.5*m*R₂²

∴    K₂ = 0.5*m*ω₂²*R₂²+0.5*(0.5*m*R₂²)*ω₂²   <em>(II)</em>

<em>∵   </em>K₁ = K₂    

⇒   0.5*m*ω₁²*R₁²+0.5*(2/3)*m*R₁²*ω₁² = 0.5*m*ω₂²*R₂²+0.5*(0.5*m*R₂²)*ω₂²

⇒  ω₁²*R₁²+(2/3)*R₁²*ω₁² = ω₂²*R₂²+0.5*R₂²*ω₂²

⇒  (5/3)*ω₁²*R₁² = (3/2)*ω₂²*R₂²

⇒  (ω₁ / ω₂)² = (3/2)*R₂² / ((5/3)*R₁²)

⇒  (ω₁ / ω₂)² = (9/10)*(7.22/ 3.59)²

⇒  (ω₁ / ω₂) = (7.22/ 3.59)√(9/10)

⇒  (ω₁ / ω₂) = 1.9079

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