Question

In: Physics

suppose that a 200g mass(0.20kg)is oscillating at the end of spring upon a frictionless horizontal surface....

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.

Solutions

Expert Solution

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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