A mass of 5.6 kg slides down a frictionless slope into a spring with spring cons

Question

A mass of 5.6 kg slides down a frictionless slope into a spring with spring constant

A mass of 5.6 kg slides down a frictionless slope into a spring with spring constant k 4.5 kN/m as depicted in the figure below. (a) If the spring experiences a maximum compression of 20 cm, what is the height h of the initial release point? (b) What is the velocity of the mass when the spring has been compressed 15 cm? (Enter the magnitude only.) (c) If the mass sticks to the end of the spring, what are the period and amplitude of the oscillations that take place? period amplitude (d) Determine the expressions for position as a function of time, x(t), and velocity as a function of time, v(t), for this oscillator. (Enter any coefficients to two decimal places. Assume x(t) is in m, and v(t) is in m/s, and that the positive direction is in the direction in which the spring is compressed x(t) v(t) (e) What is the maximum velocity of the mass during any oscillation? m/s (f) If the spring were to break while the mass is at the equilibrium position and moving to the right, to what maximum height would the mass rise up the slope? Use the exact values you enter in previous answer(s) to make later calculation(s).)

Explanation / Answer

a)

Total gravitational Potential energy stored is converted to Spring Potential energy

So, mgh = 0.5*k*x^2

So, 5.6*9.8*h = 0.5*4500*0.2^2

So, h = 1.64 m

b)

Let the KE of the body at that point = 0.5*mv^2

So, by using conservation of energy :

KE + PE,spring = PE,gravitational

So, 0.5*mv^2 + 0.5*k*x^2 = mgh

So, 0.5*5.6*v^2 + 0.5*4500*0.15^2 = 5.6*9.8*1.64

So, v = 3.75 m/s

c)

period , T = 2*pi*sqrt(m/k)

= 2*pi*sqrt(5.6/4500)

= 0.222 s

Amplitude = 20 cm = 0.2 m

d)

x(t) = 0.2*cos((2*pi/0.222)*t) = 0.2*cos(28.3*t)

v(t) = -0.2*28.3*sin(28.3*t) = -5.66*sin(28.3*t)

e)

Maximum velocity = sqrt((0.5*k*x^2)/(0.5*m))

= sqrt(4500*0.2^2/5.6) = 5.67 m/s

f)

It will rise up to the exact same height as it started (using conservation of energy)

So, h = 1.64 m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.