20. A block with a mass of 0.1 kg is attached to a spring and placed on a horizontal
frictionless table. The spring is stretched 20 cm when a force of 5 N is applied.
The spring constant is:
a. 50 Nm-1
b. 25 Nm 1
c. 75 Nm
d. 100 Nm
e. 125 Nm
Answers
0.1kg is attached to a spring and placed on a horizontal frictionless table. The spring is stretched 20cm when a force of 5N is applied. What will be the period of oscillation when the mass is set in motion?
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The spring is stretched 20 cm that is 0.20m when a force of 5 N is applied. From this information we can calculate the force constant or spring constant, k of the spring. If we denote stretching force( which is equal to restoring force in the spring by F and extension of spring by x, then
F=-kx OR
k=|F/x|=5N/0.20m=25 N/m.
Now, the spring executes simple harmonic motion on the smooth surface of the table. Then,
k=mw^2=m(4 pi^2/T^2) ……..(1).
m=0.1 kg is mass attached at the end of the spring.
T is periodic time to be found.
From equation (1),
T=2pi(m/k)^1/2. Therefore,
T=(2)(3.14)(0.1/25)^1/2 s
OR
T=0.397 s