In: Physics
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.
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