Question

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

Answer 1

Answer:

0.82052 m

Explanation:

potential energy of spring = kinetic energy

=> 0.5kx^2 = 0.5mv^2

=> $v=\sqrt{\frac{kx^2}{m} }$

$v=\sqrt{\frac{400\times 0.220^2}{2} }$

v= 3.11127 m/s

angle = 37°

thus height = distance×sin(37) = D×0.60182

Also,

m×g×h = 0.5×m×v^2

=> 2×9.8×D×0.60182 = 0.5×2×3.11127×3.11127

=> D = 0.82052 m