A 2.00-kg block is pushed against a spring with negligible mass and force consta

Question

A 2.00-kg block is pushed against a spring with negligible mass and force constant k = 400N/m , compressing it 0.220m . When the block is released, it moves along a frictionless, horizontal surface and then up a frictionless incline with slope 37.0

1)What is the speed of the block as it slides along the horizontal surface after having left the spring?

2)How far does the block travel up the incline before starting to slide back down?

A 2.00-kg block is pushed against a spring with negligible mass and force constant k = 400N/m , compressing it 0.220m . When the block is released, it moves along a frictionless, horizontal surface and then up a frictionless incline with slope 37.0 1)What is the speed of the block as it slides along the horizontal surface after having left the spring? 2)How far does the block travel up the incline before starting to slide back down?

Explanation / Answer

m = 2.00kg k = 400 N/m x = 0.220 m Energy of the compressed spring: E = .5*k*x^2 E = .5*400 N/m * (.220m)^2 E = 9.68 J Conservation of Energy, since surface is frictionless Horizontal surface (all spring energy is converted into kinetic energy) E = .5*m*v^2 v = sqrt(2E/m) v = sqrt(2*9.68J/2kg) v= 3.11 m/s Incline surface, all kinetic energy is converted to potential energy E = mgh h = E/(mg) h = 9.68 J/(2kg *9.8 m/s^2) h = 0.494m (vertical distance) along the incline is distance = h/sin(37) distance up the incline = 0.821 m

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