A 0.3 kg mass rests on top of a spring that has been compressed by 0.04 m. Neglect any frictional effects, and consider the spring to be massless. Then, if the spring has a constant k equal to 2000 N/m, to what height will the mass rise when the system is released?
a) 1.24 m
b) 0.75m
c) 0.54 m
d) 1.04 m
e) 1.34 m

20 mm

Explanation:

First, we calculate the velocity with which the object is released.

So, elastic potential energy = kinetic energy

1/2kx² = 1/2mv²

v = √(k/m)x where k = spring constant = 2000 N/m, m = mass = 0.3 kg and x = 0.04 m

v = √(2000 N/m/0.3 kg)0.04 m

v = 81.65 × 0.04 m

v = 3.27 m/s

We then calculate its acceleration, a from

F = kx = ma

a = kx/m

= 2000 N/m × 0.04 m/0.3 kg = 266.67 m/s²

We now calculate the height, h moved from v² = u² + 2ah where u = initial velocity = 3.27 m/s, v = final velocity at height, h = 0 (since it will be stationary) a = -266.67 m/s² (since its velocity is decreasing)

v² = u² + 2ah

0 =  (3.27 m/s)² + 2 ×-266.67h

-10.693 = -533.34h

h = -10.693/533.34 = 0.020 m = 20 mm

answer; ///(74) degrees is your answer..

we need to know the time interval during which the force acts. impulse of force = momentum change