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3 years ago

While eating lunch high up in a skyscraper, two construction workers calculate their gravitational potential

Physics

1 answer:

Maurinko [17]3 years ago

8 0

Answer:

The mass of the other worker is 45 kg

Explanation:

The given parameters are;

The gravitational potential energy of one construction worker = The gravitational potential energy of the other construction worker

The mass of the lighter construction worker, m₁ = 90 kg

The height level of the lighter construction worker's location = h₁

The height level of the other construction worker's location = h₂ = 2·h₁

The gravitational potential energy, P.E., is given as follows;

P.E. = m·g·h

Where;

m = The mass of the object at height

g = The acceleration due to gravity

h = The height at which is located

Let P.E.₁ represent the gravitational potential energy of one construction worker and let P.E.₂ represent the gravitational potential energy of the other construction worker

We have;

P.E.₁ = P.E.₂

Therefore;

m₁·g·h₁ = m₂·g·h₂

h₂ = 2·h₁

We have;

m₁·g·h₁ = m₂·g·2·h₁

m₁ = 2·m₂

90 kg = 2 × m₂

m₂ = (90 kg)/2 = 45 kg

The mass of the other construction worker is 45 kg.

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Three masses (3 kg, 5 kg, and 7 kg) are located in the xy-plane at the origin, (2.3 m, 0), and (0, 1.5 m), respectively.

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Answer:

a) C.M $=(\bar x, \bar y)=(0.767,0.7)m$

b) $(x_4,y_4)=(-1.917,-1.75)m$

Explanation:

The center of mass "represent the unique point in an object or system which can be used to describe the system's response to external forces and torques"

The center of mass on a two dimensional plane is defined with the following formulas:

$\bar x =\frac{\sum_{i=1}^N m_i x_i}{M}$

$\bar y =\frac{\sum_{i=1}^N m_i y_i}{M}$

Where M represent the sum of all the masses on the system.

And the center of mass C.M $=(\bar x, \bar y)$

Part a

$m_1= 3 kg, m_2=5kg,m_3=7kg$ represent the masses.

$(x_1,y_1)=(0,0),(x_2,y_2)=(2.3,0),(x_3,y_3)=(0,1.5)$ represent the coordinates for the masses with the units on meters.

So we have everything in order to find the center of mass, if we begin with the x coordinate we have:

$\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)}{3kg+5kg+7kg}=0.767m$

$\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)}{3kg+5kg+7kg}=0.7m$

C.M $=(\bar x, \bar y)=(0.767,0.7)m$

Part b

For this case we have an additional mass $m_4=6kg$ and we know that the resulting new center of mass it at the origin C.M $=(\bar x, \bar y)=(0,0)m$ and we want to find the location for this new particle. Let the coordinates for this new particle given by (a,b)