A mass of 100 g stretches a spring 5 cm. If the mass is set in motion from its equilibrium position with a downward velocity of 50 cm/s, and if there is no damping, determine the position u of the mass at any time t. (Use g = 9.8 m/s2 for the acceleration due to gravity. Let u(t), measured positive downward, denote the displacement in meters of the mass from its equilibrium position at time t seconds). What does u(t) equal?When does the mass first return to its equilibrium position? (What does t in seconds equal?)

Step-by-step explanation:

Given that:

mass m = 100 g = 0.1 kg

Length of the spring = 5 cm = 0.05 m

The set in motion from the equilibrium position u(0) = 0

The set in motion from its equilibrium position with a downward velocity u'(0) = 50 cm/s = 0.5 m/s

The spring constant (k) =

The equation of the system is expressed as:

By estimating the characteristics equation, we have r = ± 14

Thus; the general solution is:

By applying the initial condition:

u(0) = 0

⇒ 0 =

∴

u'(0) = 0.5

0.5 = 14 × c₂

c₂ = 0.5/14

c₂ = 1/28

∴

Equating u(t) = 0, we have t = π/14sec as the time when the mass first returns to its equilibrium position.

$25

step-by-step explanation:

25 - 4.50 = 20.5

20.5 / 2 = 10.25

10.25 + 10 = 20.25

20.25 - 12.50 = 7.75

therefore the answer is $25 as shown as the first number.

you probably already have the answer but anyways it's b. 15.0

step-by-step explanation: