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Bogdan [553]
3 years ago
7

Henry, whose mass is 95 kg, stands on a bathroom scale in an elevator. The scale reads 830 N for the first 3.8 s after the eleva

tor starts moving, then 930 N for the next 3.8 s. What is the elevator's speed 7.6 s after starting?
Physics
1 answer:
Delicious77 [7]3 years ago
3 0

Answer:

v= 4.0 m/s

Explanation:

  • When standing on the bathroom scale within the moving elevator, there are two forces acting on Henry's mass: Normal force and gravity.
  • Gravity is always downward, and normal force is perpendicular to the surface on which the mass is located (the bathroom scale), in upward direction.
  • Normal force, can adopt any value needed to match the acceleration of the mass, according to Newton's 2nd Law.
  • Gravity (which we call weight near the Earth's surface) can be  calculated as follows:

       F_{g} = m*g = 95 kg * 9.8 m/s2 = 930 N (1)

  • According to Newton's 2nd Law, it must be met the following condition:

       F_{net} = F_{g} -F_{n} = m*a\\  F_{net} = 930 N - 830 N = 100 N = 95 Kg * a

  • As the gravity is larger than normal force, this means that the acceleration is downward, so, we choose this direction as the positive.
  • Solving for a, we get:

       a =\frac{F_{net} }{m} =\frac{100 N}{95 kg} =  1.05 m/s2

  • We can find the speed after the first 3.8 s (assuming a is constant), applying the definition of acceleration as the rate of change of velocity:

        v_{f} = a* t = 1.05 m/s * 3.8 m/s = 4.0 m/s

  • Now, if during the next 3.8 s, normal force is 930 N (same as the weight), this means that both forces are equal each other, so net force is 0.
  • According to Newton's 2nd Law, if net force is 0, the object  is either or at rest, or moving at a constant speed.
  • As the elevator  was moving, the only choice is that it is moving at  a constant speed, the same that it had when the scale was read for the first time, i.e., 4 m/s downward.
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QUESTION 10
Elena L [17]

The maximum value of θ of such the ropes (with a maximum tension of 5,479 N) will be able to support the beam without snapping is:

\theta =37.01^{\circ}

We can apply the first Newton's law in x and y-direction.

If we do a free body diagram of the system we will have:

x-direction

All the forces acting in this direction are:

T_{1}sin(\theta)-T_{2}sin(\theta)=0    (1)

Where:

  • T(1) is the tension due to the rope 1
  • T(2) is the tension due to the rope 2

Here we just conclude that T(1) = T(2)

y-direction

The forces in this direction are:

T_{1}cos(\theta)+T_{2}cos(\theta)-W=0   (2)

Here W is the weight of the steel beam.

We equal it to zero because we need to find the maximum angle at which the ropes will be able to support the beam without snapping.

Knowing that T(1) = T(2) and W = mg, we have:

T_{1}cos(\theta)+T_{1}cos(\theta)-m_{steel}g=0

2T_{1}cos(\theta)-m_{steel}g=0

2T_{1}cos(\theta)=m_{steel}g

T(1) must be equal to 5479 N, so we have:

cos(\theta)=\frac{m_{steel}g}{2T_{1}}

cos(\theta)=\frac{892*9.81}{2*5479}

cos(\theta)=\frac{892*9.81}{2*5479}

cos(\theta)=0.80

Therefore, the maximum angle allowed is θ = 37.01°.

You can learn more about tension here:

brainly.com/question/12797227

I hope it helps you!

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