An object with a mass of 200 g rests on a coiled spring with a spring constant of k=500 N/m. The spring is compressed by 5 cm.
a. What is the energy stored in the spring? b. How much energy would the spring give to the object?
c. How high will the object reach?
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Ответ:
Velocity will be![v(t)=\frac{3t^2}{2}+5t-4ft/sec](/tpl/images/0255/7915/3f69e.png)
Distance traveled = 6 feet
Explanation:
We have given acceleration![a(t)=3t+5](/tpl/images/0255/7915/53b93.png)
We know that![v(t)=\int a(t)dt](/tpl/images/0255/7915/d57ee.png)
So![v(t)=\int 3t+5dt](/tpl/images/0255/7915/e1a95.png)
We have given![v(0)=-4ft/sec](/tpl/images/0255/7915/4e94a.png)
So![-4=\frac{3\times 0^2}{2}+5\times 0+c](/tpl/images/0255/7915/9c60f.png)
c = -4 ft/sec
So![v(t)=\frac{3t^2}{2}+5t-4ft/sec](/tpl/images/0255/7915/3f69e.png)
Now we have to find distance traveled in interval [0,2]
So distance![s=\int v(t)dt](/tpl/images/0255/7915/ef874.png)