suppose that a 200g mass(0.20kg)is oscillating at the end of spring upon a frict
ID: 1324196 • Letter: S
Question
suppose that a 200g mass(0.20kg)is oscillating at the end of spring upon a frictionless horizontal surface. The spring can be both stretched and compressed and has a spring constant of 240N/m. It was originally stretchef a distance of 12m (0.12m) from it equilibrium.
What is its initial potential energy (PEmax) show calculation
where in the oscillation is this maximum reached?
what is the maximum velocity (Vmax) that the mass will reach in its oscillation? show calculation
when the mass is 6cm from the equilibrium position, what are the values of the elastic potential energy
PE1 equal to
Kinetic energy KE1 is equal to
Velocity V1 is equal to
Kindly provide detailed step by step solution for each question.
Explanation / Answer
The initial potential energy is
U = ( 1/ 2) k x ^2 = ( 1/2 ) ( 240 N / m ) ( 0.12 m )^2
= 1.728 J
Maximum velocity of the mass is
v = sqrt ( k / m ) x = sqrt( ( 240 N /m) / ( 0.20 kg ) ) (0.12 m) = 4.156 m/s
potential energy when the mass is at 6 cm is
U' = ( 1/ 2) ( 250 N /m ) ( 0.06 m )^2 = 0.45 J
kinetic energy
K = ( 1/2 ) k A^2 - ( 1/ 2 ) k x ^2
= ( 0.5 ) ( 250 N /m ) ( 0.12 ^2 - 0.06 ^2 )
= 1.35 J
velocity of the mass is
v = sqrt ( 2 K / m ) = sqrt ( 2 ( 1.35 J / 0.2 kg ))
= 3.67 m/s
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