A 2.00 kg package is released on a 53.1 degree incline, 4.00m from a long spring

Question

A 2.00 kg package is released on a 53.1 degree incline, 4.00m from a long spring with force constant 120 N/m that is attached at the bottom of the incline . The coefficients of friction between the package and the incline are µs= 0.40 and µk=0.20. The mass of the spring is negligible. The speed of the package just before it hits the spring is 7.30 m/s (that was part a and I found that out) What is the maximum compression of the spring? A 2.00 kg package is released on a 53.1 degree incline, 4.00m from a long spring with force constant 120 N/m that is attached at the bottom of the incline . The coefficients of friction between the package and the incline are µs= 0.40 and µk=0.20. The mass of the spring is negligible. The speed of the package just before it hits the spring is 7.30 m/s (that was part a and I found that out) What is the maximum compression of the spring? What is the maximum compression of the spring?

Explanation / Answer

Let maximum compression of the spring be x Comparing energy at the point when block just touches spring and at the maximum compression point. Internal Energy = energy stored in spring + kinetic energy of package + potential energy of package Initially, energy stored in spring is zero At maximum compression, velocity of package and hence kinetic energy will be zero. Additionally it will lose energy for the work done against friction. So, Initial energy - Final energy = work done against friction work done against friction = force x distance =(umgcos?)(x) (1/2)mv2 + mgh - (1/2)kx2 =umgcos(?)x Where h is the vertical distance moved = xsin(?) 60x2 - 13.31x - 53.25 = 0 so, x = 1.06 m

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