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Radda [10]
1 year ago
8

A 0.20 kg mass attached to the end of a spring causes it to stretch 3.0 cm. What is the spring constant? What is the potential e

nergy of the spring?
Physics
1 answer:
SVEN [57.7K]1 year ago
8 0

Given that the mass is m = 0.2 kg and the displacement is x = 3 cm = 0.03 m

We have to find the spring constant and potential energy.

The spring constant can be calculated by the formula

\begin{gathered} F=\text{ kx} \\ mg\text{ = kx} \\ k\text{ = }\frac{mg}{x} \end{gathered}

Here, k is the spring constant.

g = 9.8 m/s^2 is the acceleration due to gravity.

Substituting the values, the spring constant will be

\begin{gathered} k=\frac{0.2\times9.8}{0.03} \\ =\text{ 65.33 N/m } \end{gathered}

The potential energy can be calculated as

\begin{gathered} U=\frac{1}{2}kx^2 \\ =\frac{1}{2}\times65.33\text{ }\times(0.03)^2 \\ =\text{ 0.0293 J} \end{gathered}

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A baseball player leads off the game and hits a long home run. The ball leaves the bat at an angle of 70.0 from the horizontal w
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Answer: 211.059 m

Explanation:

We have the following data:

\theta=70\° The angle at which the ball leaves the bat

V_{o}=55 m/s The initial velocity of the ball

g=-9.8 m/s^{2} The acceleration due gravity

We need to find how far (horizontally) the ball travels in the air: x

Firstly we need to know this velocity has two components:

<u>Horizontally:</u>

V_{ox}=V_{o}cos \theta (1)

V_{ox}=55 m/s cos(70\°)=18.811 m/s (2)

<u>Vertically:</u>

V_{oy}=V_{o}sin \theta (3)

V_{oy}=55 m/s sin(70\°)=51.683 m/s (4)

On the other hand, when we talk about parabolic movement (as in this situation) the ball reaches its maximum height just in the middle of this parabola, when V=0 and the time t is half the time it takes the complete parabolic path.

So, if we use the following equation, we will find t:

V=V_{o}+gt=0 (5)

Isolating t:

t=\frac{-V_{o}}{g} (6)

t=\frac{-55 m/s}{-9.8 m/s^{2}} (7)

t=5.61 s (8)

Now that we have the time it takes to the ball to travel half of is path, we can find the total time T it takes the complete parabolic path, which is twice t:

T=2t=2(5.61 s)=11.22 s (9)

With this result in mind, we can finally calculate how far the ball travels in the air:

x=V_{ox}T (10)

Substituting (2) and (9) in (10):

x=(18.811 m/s)(11.22 s) (11)

Finally:

x=211.059 m

8 0
3 years ago
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