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Nata [24]
2 years ago
8

In a "worst-case" design scenario, a 2000-kg elevator with broken cables is falling at 4.00 m/s when it first contacts a cushion

ing spring at the bottom of the shaft. The spring is supposed to stop the elevator, compressing 2.00 m as it does so. During the motion a safety clamp applies a constant 17000-N frictional force to the elevator. 1) What is the speed of the elevator after it has moved downward 1.00 m from the point where it first contacts a spring? 2) When the elevator is 1.00 m below point where it first contacts a spring, what is its acceleration?
Physics
1 answer:
Fofino [41]2 years ago
3 0

Answer:

Part a)

v_f = 3.65 m/s

Part b)

a = - 4 m/s^2

Explanation:

Part a)

As we know that the elevator is stopped while spring is compressed by x = 2 m

now we will have

\frac{1}{2}kx^2 + F_f x - mgx = \frac{1}{2}mv^2

\frac{1}{2}k(2^2) + (17000)(2) - (2000\times 9.81\times 2) = \frac{1}{2}(2000)(4^2)

2k + 34000 - 39240 = 16000

k = 10620 N/m

Now we have to find the speed when spring is compressed by x = 1

So again by work energy theorem

W_g + W_f + W_{spring} = \frac{1}{2}m(v_f^2 - v_i^2)

2000(9.81)(1) - 17000(1) - \frac{1}{2}(10620)(1^2) = \frac{1}{2}(2000)(v_f^2 - 4^2)

-2.69 = v_f^2 - 16

v_f = 3.65 m/s

Part b)

Net force on the elevator while spring is compressed by x = 1

F_{net} = mg - kx - F_f

F_{net} = 2000(9.81) - (10620\times 1) - 17000

F_{net} = -8000 N

now acceleration is given as

a = \frac{F}{m}

a = -\frac{8000}{2000}

a = - 4 m/s^2

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Explanation:

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2 years ago
A 7 kg ball is moving at a constant speed of 5 m/s. A force of 300 N is applied to the ball for 4 s. The new speed of the ball i
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The new speed of the ball is 176.43 m/s

Explanation:

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Apply Newton's second law of motion;

F = ma = \frac{m(v-u)}{t}\\\\m(v-u) = Ft\\\\v-u = \frac{Ft}{m}\\\\v =  \frac{Ft}{m} + u

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A wire 2.80 m in length carries a current of 5.60 A in a region where a uniform magnetic field has a magnitude of 0.300 T. Calcu
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Complete question:

A wire 2.80 m in length carries a current of 5.60 A in a region where a uniform magnetic field has a magnitude of 0.300 T. Calculate the magnitude of the magnetic force on the wire assuming the following angles between the magnetic field and the current.

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c) 120 ⁰

Answer:

(a) When the angle, θ = 60 ⁰,  force = 4.07 N

(b) When the angle, θ = 90 ⁰,  force = 4.7 N

(c) When the angle, θ = 120 ⁰,  force = 4.07 N

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Given;

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Answer:

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