Answer:
Spring Constant = 279.58 N/m
Explanation:
We are given;
Mass; m = 2.05 x 10^(-2) kg = 0.0205 kg
Distance of compression; x = 8.01 × 10^(-2) m = 0.0801 m
Maximum height; h = 4.46 m
The formula for the energy in the spring is given by;
E = ½kx²
where:
k is the spring constant
x is the distance the spring is compressed.
Now, this energy of the spring will be equal to the energy of the pellet at its highest point. Energy of pallet = mgh So;
½kx² = mgh
Plugging in the relevant values, we have;
½ * k * 0.0801² = 0.0205 * 9.81 * 4.46
0.003208005k = 0.8969
k = 0.8969/0.003208005
k = 279.58 N/m
Answer:
a)
Nm⁻¹
b)
Hz
c)
ms⁻¹
d)
m
e)
ms⁻²
f)
m
Explanation:
a)
= force required to hold the object at rest connected with stretched spring = 27 N
= stretch in the spring from equilibrium position = 0.2 m
= force constant of the spring
force required to hold the object at rest is same as the spring force , hence
inserting the values
Nm⁻¹
b)
frequency of the oscillations is given as
inserting the values
Hz
c)
= Amplitude of oscillations = 0.2 m
= angular frequency
Angular frequency is given as
rads⁻¹
Maximum speed of oscillation is given as
ms⁻¹
d)
maximum speed of the object occurs at the equilibrium position, hence
m
e)
Maximum acceleration of oscillation is given as
ms⁻²
f)
maximum acceleration occurs when the object is at extreme positions, hence
m
