A 0.15kg block is connected to a spring with spring constant k= 60 N/m on a hori

Question

A 0.15kg block is connected to a spring with spring constant k= 60 N/m on a horizontal frictionless table. You compress the spring a distance of 0.5m and let the block go.

a) what is the angular frequency of the oscillation?

b) What is the mechanical enegry of the block-spring system?

c) What are the position, speed, potential and kinetic enerdies of the block-spring system when: 1.the block is at a turning point? 2.the pring is at its equilibrium length? 3. the spring is compressed 0.3m? 4. the block is moving at a speed of 6m/s?

d) When the spring is stretched by 0.4m, the block breaks free and slides off the edges of the 0.8m tall table. How fast is the block going when it hits the floor?

Explanation / Answer

Solution :

mass , m= 0.15 kg , Spring constant K = 60 N/m
amplitude A = 0.5 m
a) w = sqrt(k/m) = 20 rad/s
b)E =0.5kA^2 = 0.5*60*0.5^2 = 7.5 J
d) 0.5kx^2 +mgh = 0.5mv^2

0.5*60*0.4^2 + (0.15*9.8*0.8) = 0.5*0.15*v^2

speed v = 8.926 m/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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