At an outdoor market, a bunch of bananas is set into oscillatory motion with an amplitude of 29.8981 cm on a spring with a spring constant of 19.1461 N/m. The mass of the bananas is 43.288 kg. What is the maximum speed of the bananas

Explanation:

Given that,

The amplitude of oscillation is

A = 29.8981cm = 0.298981m

Spring constant k = 19.1461N/m

Mass of banana hung m = 43.288kg

Maximum speed v?

The maximum speed can be determined by using the formula

v = w•A

Where

w = angular frequency

A = amplitude

So we need to get the angular frequency and it can be determined by using the formula

w = √ (k/m)

w = √(19.1461/43.288)

w = √0.4423

w = 0.6651rad/sec

Then, maximum speed

v = w•A

v = 0.6651 × 0.298981

v = 0.19884m/s

v ≈ 0.2m/s

The maximum speed of oscillation is 0.2m/s

their crystal patterns have been disrupted by stress

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