An object is dropped from a vertical height of 1.89 m above the balcony level. What is the object’s speed when it is 2.20 m below the balcony level if 10.0% energy is lost due to the air resistance? Does it matter when to apply 10% loss before V calculations or after? [8.49m/s] [yes it does, 0.9Energy result in √0.9Velocity]

Answers

Answer 1

a.

The object's speed at 2.20 m below balcony level is 8.74 m/s

Let the balcony level be 0 m and the height above the balcony level be positive and height below the balcony level negative.

Using the principle of conservation of energy, the total energy at a vertical height of 1.89 m above the balcony level equals the total mechanical energy when the object is 2.20 m below the balcony level and

So, E = E'

U + K + f = U' + K' + f'

where U = initial potential energy at 1.89 m = mgh, K = initial kinetic energy at 1.89 m = 0 J(since it is released from rest), f = energy loss at 1.89 m = 0 J, U' = final potential energy at 2.20 m below balcony level = mgh', K = final kinetic energy at 2.20 m = 1/2mv², f' = energy loss at 1.89 m = 10%U = 0.10mgh(since 10% of the initial energy is lost).

So,

U + K + f = U' + K' + f'

mgh + 0 + 0 = mgh' + 1/2mv² + 0.10mgh

mgh = mgh' + 1/2mv² + 0.10mgh

Dividing through by m, we have

gh = gh' + 1/2v² + 0.10gh

So, gh -  0.10gh = gh' + 1/2v²

0.90gh = gh' + 1/2v²

1/2v² = 0.90gh - gh'

1/2v² = g(0.90h - h')

v² = 2g(0.90h - h')

Taking square-root of both sides, we have

v = √[2g(0.90h - h')]

where v = velocity of object at 2.20 m below balcony level, h = height above the balcony level = 1.89 m, h' = height below the balcony level = -2.20 m and  g = acceleration due to gravity = 9.8 m/s²

Substituting the values of the variables into the equation, we have

v = √[2g(0.90h - h')]

v = √[2 × 9.8 m/s²{0.90 × 1.89 m - (-2.20 m)}]

v = √[2 × 9.8 m/s²(1.701 m + 2.20 m)]

v = √[2 × 9.8 m/s²(3.901 m)]

v = √[76.4596 m²/s²]

v = 8.74 m/s

So, the object's speed at 2.20 m below balcony level is 8.74 m/s

b.

Yes it does matter when we apply 10% loss before V calculations

We need to apply the 10 % loss before V calculations because this would give us a proper value for V since the energy is lost before V is obtained.

So, yes it does matter when we apply 10% loss before V calculations

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ANSWER:

What is the acceleration of the cart at t=8 seconds?

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