hadisheeler378
contestada

7 A 0.25 kg block oscillates on the end of a spring with a spring constant of 100 N/m. If the oscillation is started by elongating the spring 0.1m and giving the block a speed of 3 mis then the amplitude of the oscillation is:​

Respuesta :

toreehines18
(K+U)
i
​
=(K+U)
f
​


0+
2
1
​
kA
2
=
2
1
​
mv
2
+
2
1
​
kx
2



2
1
​
(6.50 N/m)(0.100 m)
2
=
2
1
​
m(0.300 m/s)
2
+
2
1
​
(6.50 N/m)(5.00×10
−2
m)
2


3.25×10
−2
J=
2
1
​
m(0.300 m/s)
2
+8.12×10
−3
J

giving m=
9.0×10
−2
m
2
/s
2

2(2.44×10
−2
J)
​
=
0.542 kg
​


(b) ω=
m
k
​

​
=
0.542 kg
6.50 N/m
​

​
=3.46 rad/s

Then, T=
ω
2π
​
=
3.46 rad/s
2π rad
​
=
1.81 s
​


(c) a
max
​
=Aω
2
=(0.100 m)(3.46 rad/s)
2
=
1.20 m/s
2

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