Answer:
0.466 m/s
Explanation:
We are given that
Mass =m=0.46 kg
Spring constant,k=39.9 N/m
Length of stretched spring=A=5 cm=[tex]5\times 10^{-2} m[/tex]
1 m=100 cm
a.We have to find the maximum speed of the oscillating mass.
Maximum speed,v=[tex]A\sqrt{k}{m}}[/tex]
Using the formula
Maximum speed,v=[tex]5\times 10^{-2}\sqrt{\frac{39.9}{0.46}}[/tex]
Maximum speed,v=0.466 m/s
Answer:
Explanation:
mass of block, m = 0.46 kg
Spring constant, K = 39.9 N/m
extension in spring, Δs = 5 cm = 0.05 m
By the conservation of energy, the maximum potential energy stored in the spring is equal to the kinetic energy of the block.
1/2 K Δs² = 1/2 m v²
39.9 x 0.05 x 0.05 = 0.46 x v²
v = 0.47 ms
