A block weighing 14.0 N is attached to the lower end of a vertical spring (k = 2

Question

A block weighing 14.0 N is attached to the lower end of a vertical spring (k = 238.0 N/m), the other end of which is attached to a ceiling. The block oscillates vertically and has a kinetic energy of 1.90 J as it passes through the point at which the spring is unstretched. (a) What is the period of the oscillation? (b) Use the law of conservation of energy to determine the maximum distance the block moves above the point at which the spring is unstretched. (The maximum points above and below are not necessarily the same.) (c) Use the law of conservation of energy to determine the maximum distance the block moves below the point at which the spring is unstretched. (d) What is the amplitude of the oscillation? (e) What is the maximum kinetic energy of the block as it oscillates?

Explanation / Answer

(a) w=sqrt(k/m)=12.9

period=2*pi/w=0.486 sec

(b)KE=0.5*mv^2

we got v=sqrt(2*KE/m)=1.63 m/s

Since oscillattion is SHM

let

x=Asin(wt+B)

velocity=v=dx/dt=Awcos(wt+B)

at unstretched position,v=wA

let it is stretched x m upward from unstretched position

conserving Energy

0.5*m*v^2=0.5*k*x^2+(mgx)

we got x=0.081=8.1 cm

(c)Let it go x m down

Conserving Energy

1.9=119x^2-14x

x=0.198 m=19.8 cm

(d)Amplitude=19.8 cm

(e)At any point let kinetic energy is KE

1.9=KE+0.5kx^2+mgx..............(i)

KE=1.9+mgx-0.5kx^2

for maximum .dKE/dx=0

we got x=0.058 m

putting this x in equation(i)

maximum kinetic energy=2.3 J

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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