4. A massless spring hangs from the ceiling, and a mass is hung from the bottom of it. The mass is supported so that initially the tension in the spring is zero. The mass is then suddenly released. At the bottom of
its trajectory, the mass is 5 centimeters from its original position. Find its oscillation period.
(A) 0.05 s
(B) 0.07 s
(C) 0.31 s
(D) 0.44 s
(E) Not enough information is given.

The correct option is C: 0.31 s.

Explanation:

When the mass is then suddenly released we have:

Where:

F is the force

k: is the spring constant

Δy: is the spring displacement

Since the tension in the spring is zero, the force is the weight:

Where:

m is the mass of the object

g is the gravity

   (1)

The oscillation period of the spring is given by:

   (2)

By solving equation (1) for "k" and entering into equation (2) we have:

Since the spring will osclliates in a position between the initial position (when it is at rest) and the final position (when the mass is released and reaches the bottom), we have Δy = 2.5 cm = 0.025 m:

Hence, the oscillation period is 0.31 s.

The correct option is C: 0.31s.

I hope it helps you!                                                                                                      

v=ir = 3x15 =45 v

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards