A 0.400 kg mass is attached to a horizontal spring with a spring constant of 50.
ID: 1426487 • Letter: A
Question
A 0.400 kg mass is attached to a horizontal spring with a spring constant of 50.0 N/m. The mass is initially stretched a distance of 48.0 cm from its equilibrium position. The mass is then released from rest and slides along a horizontal table.
A) Assume the table is frictionless and there is no air resistance. Select the energy types which change from the time the mass is released to the time it is compressed a distance of 39.0 cm on the other side of equilibrium. K , Ug, Uel, Uint
B) What will be the speed of the mass when it is compressed to the distance of 39.0 cm from equilibrium?
Now we will imagine that the table does exert a kinetic friction force on the mass (we will still ignore air resistance, however). We would like to analyze this scenario in two ways: We will treat the mass-spring as an open system, where the table is exerting an external force on the mass. This system will be analyzed in parts C, D and E. We will treat the mass-spring and the table as a closed system, with no external forces acting. This system will be analyzed in parts F and G.
C) Due to the kinetic friction from the table, after being released as before (from rest, stretched a distance of 48.0 cm from equilibrium), the mass reaches its maximum compression 40.0 cm on the other side of equilibrium. Find the change in the mechanical energy of the mass-spring from when it is released to when it reaches its maximum compression. (Remember that mechanical energy is defined as the sum of the kinetic energy and total potential energy of the system).
D) Using your answer to part C, find the work done by the external force of the table on the mass
E) Use your answer to part D to solve for the coefficient of kinetic friction between the table and the mass.
F) In this system, select which types of energy must change from the initial state (released from rest stretched 48 cm from equilibrium) to the final state (maximum compression 40 cm from equilibrium)? Ug, Uel, Uint, K
G) Use conservation of energy to solve for the change in the internal energy of the system.
Explanation / Answer
A) Initially there was only potential energy but later there will be only kinetic energy when mass is at equiliberium position. then kinetic energy starts decreasing and at 39 cm compression it will have some kinetic energy and some potential energy.
B) KE_F + PE_F = KE_I + PE_I
PE = 0.5*k*x^2
KE = 0.5*m*v^2
=> KE_F = 0.5*50*(0.48^2 - 0.39^2) = 1.9575 = 0.5*0.4*v^2
=> v = sqrt(1.9575*2/0.4) = 3.128 m/s Answer
C) Change in ME = Initial PE - Final PE = 0.5*50*0.48^2 - 0.5*50*0.4^2 = 1.76 J Answer
D) This change in ME is due to work done by enternal force. So Work = 1.76 J Answer
E) Friction force(f) = mu_k*mg
Work = F.s = mu_k*mg*(0.48+0.4) = 1.76
=> mu_k = 1.76/(0.4*9.81*(0.48+0.4)) = 0.509 Answer
F) Potential Energy
G) Change = 1.76 J decrease.
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