A 2-kg block is pressed into a spring with constant k = 1450 N/m. The spring is

Question

A 2-kg block is pressed into a spring with constant k = 1450 N/m. The spring is compressed 25 cm from its natural position. The block is released and moves along the path shown below. The path's surface from the position of the spring to the right-hand side of the bump is frictionless. What is the speed of the block after the spring launch? What is the speed of the block when it reaches the top of the 2.0 m bump (2.0 meter above the starting position)? What is the speed of the block when it reaches the right-hand side of the bump (1.0 meter above the starting position)?

Explanation / Answer

using the conservation of energy,
0.5kx2 = 0.5mu^2

1450( 0.25)^2 = 2(u)^2

solvoing for u

u = 6.73 m/s apprx ( velocity after the sping launch)

b) 0.5m(6.73)^2 = mgh + 0.5mV2

solving for V ( velocity at the height of 2 m)

22.656 = 9.8(2) + 0.5(v)2

V = 2. 47 m/s apprx ( velocity at top of 2m)

c)  0.5m(6.73)^2 = mgh + 0.5mV2 ( where V is now velocity at the top of 1 m)

22.656 = 9.8(1) + 0.5v^2

v = 5.07 m/s apprx ( velocity at the height of 1 m)

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