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4 years ago

15

A 3.5 kg block slides along a frictionless horizontal with a speed of 2.5 m/s. After sliding a distance of 5 m, the block makes

a smooth transition to a frictionless ramp inclined at an angle of 50â—¦ to the horizontal. The acceleration of gravity is 9.81 m/s 2 .

1 answer:

Andru [333]4 years ago

3 0

Answer:

L = 0.416 m

Explanation:

In this exercise to calculate the distance traveled in the branch that climbs the block we can use the concepts of energy

Starting point Flat part

Em₀ = K = ½ m v²

Final point. Highest part of the ramp

$Em_{f}$ = U = m g h

As there is no friction the mechanical energy is conserved

Em₀ = $Em_{f}$

½ m v² = m g h

h = v² / 2g

h = 2.5 2/2 9.8

h = 0.3189 m

The distance traveled can be found with trigonometry

sin 50 = h / L

L = h / sin 50

L = 0.3189 / sin50

L = 0.416 m

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3 years ago

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$\= F = 697*10^{-15}(-j) N$

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<em />

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(1 kg) <em>g</em> sin(45°) - (0.4 kg) <em>g</em> = (1.4 kg) <em>a</em>

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8 0

4 years ago