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sweet [91]
3 years ago
13

You've always wondered about the acceleration of the elevators in the 101 story-tall Empire State Building. One day, while visit

ing New York, you take your bathroom scale into the elevator and stand on them. The scales read 160 lb as the door closes. The reading varies between 130 lb and 180 lb as the elevator travels 101 floors.
a. What is the magnitude of the acceleration as the elevator starts upward?
b. What is the magnitude of the acceleration as the elevator brakes to a stop?
Physics
1 answer:
Arte-miy333 [17]3 years ago
8 0

Answer:

a)  a = 4 ft / s² , b) a = -6 ft / s²

Explanation:

The balance is subjected to two forces: the weight of the person directed downward and the spring reaction directed upward.

When the person rides the elevator, the acceleration is zero

            F - W = 0

            F = W

            F = 160 lb

let's find the mass of the body

            W = mg

            m = W / g

            m = 160/32

           m = 5 slug

A) when the elevator is moving up

            F - W = m a

            F = W + m a

            F - W / m = m a

            F = m (g + a)

therefore the scale reading (F) must be higher, in this case F = 180 lb

             a = F / m - g

            a = 180 - 160)/5

            a = 4 ft / s²

b) when the elevator is stopping

in this case the direction is opposite to the speed, that is to say downwards

              F- W = m (-a)

              a = W - F / m

              a = 130 -160 /5

              a = -6 ft / s²

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A 210 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 2.0 N/cm. The block becomes attached t
Yuliya22 [10]

Answer:

a) W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J

b) W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J

c) V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s

d)  d_1 =0.183m or 18.3 cm

Explanation:

For this case we have the following system with the forces on the figure attached.

We know that the spring compresses a total distance of x=0.10 m

Part a

The gravitational force is defined as mg so on this case the work donde by the gravity is:

W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J

Part b

For this case first we can convert the spring constant to N/m like this:

2 \frac{N}{cm} \frac{100cm}{1m}=200 \frac{N}{m}

And the work donde by the spring on this case is given by:

W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J

Part c

We can assume that the initial velocity for the block is Vi and is at rest from the end of the movement. If we use balance of energy we got:

W_{g} +W_{spring} = K_{f} -K_{i}=0- \frac{1}{2} m v^2_i

And if we solve for the initial velocity we got:

V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s

Part d

Let d1 represent the new maximum distance, in order to find it we know that :

-1/2mV^2_i = W_g + W_{spring}

And replacing we got:

-1/2mV^2_i =mg d_1 -1/2 k d^2_1

And we can put the terms like this:

\frac{1}{2} k d^2_1 -mg d_1 -1/2 m V^2_i =0

If we multiply all the equation by 2 we got:

k d^2_1 -2 mg d_1 -m V^2_i =0

Now we can replace the values and we got:

200N/m d^2_1 -0.21kg(9.8m/s^2)d_1 -0.21 kg(5.50 m/s)^2) =0

200 d^2_1 -2.058 d_1 -6.3525=0

And solving the quadratic equation we got that the solution for d_1 =0.183m or 18.3 cm because the negative solution not make sense.

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Answer:

W=1705.2 J

Explanation:

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mass ,m= 60 kg

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Height ,h= 2.9 m

As we know that work done by a force given as

W = F . d

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d=Displacement

W=work done by force

Now by putting the values

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W=1705.2 J

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