A spring of negligible mass stretches 3.00 cm from its relaxed length when a for

Question

A spring of negligible mass stretches 3.00 cm from its relaxed length when a force of 7.40 N is applied. A 0.510-kg particle rests on a frictionless horizontal surface and is attached to the free end of the spring. The particle is displaced from the origin to x = 5.00 cm and released from rest at t = 0. (Assume that the direction of the initial displacement is positive.)

(g) Determine the velocity and acceleration of the particle when t = 0.500 s. (Indicate the direction with the sign of your answer.)

Explanation / Answer

now spring constant K=7.4/0.03

=246.67

Now w=sqrt(k/m)

=22 s-1

TIme period=0.3 s

Now at t=0.5 it will be at 3.33 cm and moving away from the spring

velocity can be calculated on the basis of conservation of energy

1/2*m*v^2=1/2*k*(x2^2-x1^2)

v=0.82 m/s

a=-w^2*x

=-22*3.33*10^-2

= -0.73 m/s^2 ans

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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