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aleksley [76]
3 years ago
10

A cat’s crinkle ball toy of mass 15g is thrown straight up with an initial speed of 3m/s. Assume in this problem that air drag i

s negligible. If the gravitational potential energy is taken to be zero at the point where it leaves your hand, what is the gravitational potential energy when the ball is at its peak height?
Physics
1 answer:
stiks02 [169]3 years ago
4 0

Answer:

P.E = 0.068 J = 68 mJ

Explanation:

First we need to find the height attained by the ball toy. For this purpose, we will be using 3rd equation of motion:

2gh = Vf² - Vi²

where,

g = -9.8 m/s²  (negative sign due to upward motion)

h = height attained by the ball toy = ?

Vf = Final Velocity = 0 m/s (since it momentarily stops at the highest point)

Vi = Initial Velocity = 3 m/s

Therefore,

2(-9.8 m/s²)h = (0 m/s)² - (3 m/s)²

h = (9 m²/s²)/(19.6 m/s²)

h = 0.46 m

Now, the gravitational potential energy of ball at its peak is given by the following formula:

P.E = mgh

P.E = (0.015 kg)(9.8 m/s²)(0.46 m)

<u>P.E = 0.068 J = 68 mJ</u>

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Answer:

a

 k =  457333.3 N/m

b

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Explanation:

From the question we are told that

    The total mass of  three people is  M  = 2.00*10^{2} \ kg

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Generally the spring constant is mathematically represented as

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Here F is the force exerted by the mass of three people and that of the car , this is mathematically represented as        

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So

        k =  \frac{13720}{0.03}

=>     k =  457333.3 N/m

Generally if the mass which the car is loaded with is  m  =  3.00*10^{2} \ kg

Then the force experienced by the spring is  

         =>       F_a = (m +m_c) *g

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       x_a  = \frac{F_a}{k}

=>    x_a  = \frac{41160}{457333.3}

=>    x_a  =0.09\ m      

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