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Harrizon [31]
3 years ago
14

A person carries a plank of wood 2 m long with one hand pushing down on it at one end with a force F1 and the other hand holding

it up at 50 cm from the end of the plank with force F2 . If the plank has a mass of 20 kg and its center of gravity is at the middle of the plank, what are the magnitudes of the forces F1 and F2 ?

Physics
1 answer:
sveta [45]3 years ago
5 0

Answer:

F1= 588 N

F2= 784 N

Explanation:

Please see the attached file.

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What speed would a fly with the mass of 0.55 g need in order to have the same kinetic energy as the automobile in the term 19
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<u>The question does not provide enough information to complete the answer, so I'll assume the needed data to help you to solve your own problem</u>

Answer:

<em>The fly should need to move at 9,534.6 m/s to have the same kinetic energy as the automobile</em>

Explanation:

<u>Kinetic Energy </u>

Is the capacity of a body to do work due to its speed and is computed by

\displaystyle K=\frac{mv^2}{2}

We are not given enough data to compare the kinetic energy of the fly with that of the automobile. We'll assume the following characteristics:  

m_a=500\ kg

v_a=10\ m/s

So its kinetic energy is

\displaystyle K_a=\frac{(500)10^2}{2}

\displaystyle K_a=25,000\ J

The mass of the fly is  

m_f=0.55\ gr=0.00055\ kg

To have the same kinetic as the automobile:

\displaystyle \frac{m_fv_f^2}{2}=25,000

Solving for v_f

\displaystyle v_f=\sqrt{\frac{2(25,000))}{m_f}}

\displaystyle v_f=\sqrt{\frac{50,000}{0.00055}}

v_f=9,534.6\ m/s

The fly should need to move at 9,534.6 m/s to have the same kinetic energy as the automobile

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Determine the magnitude of the average friction force exerted on the collar when the velocity of the collar at c is 3.39 m/s and
djverab [1.8K]

The magnitude of the average friction force exerted on the collar (F)=  8.641 N

<h3>How can we calculate the magnitude of the average friction force exerted on the collar?</h3>

To calculate the magnitude of the average friction force exerted on the collar we are using the formula,

\frac{1}{2} k(x^2_f - x^2_i ) + F\times y + \frac{1}{2} m v^{2} _{c}  = mgy

Here we are given,

k = The spring has a spring constant.

= 25.5 N/m.

x_f = Final length of the spring .

= \sqrt{1.25^2+1.8^2}  -0.60

= 1.591 m

x_i= The initial length of the spring.

= 1.25−0.60

=0.65 m

y=The collar then travels downward a distance.

=  1.80 m.

m= The mass of the collar.

=3.55 kg

v_c = the velocity of the collar.

= 3.39 m/s.

g = The acceleration due to gravity.

= 9.81 m/s²

We have to calculate the magnitude of the average friction force exerted on the collar = F

Now we put the known values in the above equation, we get;

\frac{1}{2} k(x^2_f - x^2_i ) + F\times y + \frac{1}{2} m v^{2} _{c}  = mgy

Or, \frac{1}{2} \times 25.5 \times((1.591)^2 - (0.65)^2 ) + F\times 1.80 + \frac{1}{2}\times 3.55\times (3.39)^{2}  = 3.55\times 9.81\times 1.80

Or, F= 8.641 N

From the above calculation we can conclude that,

The magnitude of the average friction force exerted on the collar (F)=  8.641 N

Learn more about friction:

brainly.com/question/24338873

#SPJ4

Disclaimer: This question is incomplete in the portal. Here is the complete question.

Question:

The 3.55 kg collar shown below is attached to a spring and released from rest at A. The collar then travels downward a distance of y = 1.80 m. The spring has a spring constant of k = 25.5 N/m. The distance a is given as 1.25 m. The datum for gravitational potential energy is set at the horizontal line through A and B. Determine the magnitude of the average friction force exerted on the collar when the velocity of the collar at c is 3.39 m/s and the spring has an unstretched length of 0.60 m .

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2 years ago
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