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------------------------------------------------ A horizontal spring-mass system has low friction, a spring stiffness of 215 N/m, and a mass of 0.3 kg. The system is released with an initial compression of the spring of 8 cm and an initial speed of the mass of 3 m/s.

What is the maximum stretch during the motion?

Answer :

Answer:(a) the maximum stretch of the spring is 17.7cm.

(b) the maximum speed during the motion is 4.08 m/s.

(c) the average power input required is 0.11W.

Spring-mass system:

(a) The law of conservation of energy suggests that the total mechanical energy of the system remains conserved in case of a constant or conservative force. So,

the initial energy of the spring-mass system = final energy

also, the total energy remains conserved, so, when the potential energy is maximum, the kinetic energy is zero.

where, u = 3 m/s is the initial speed of the mass m = 0.3kg

k = 160 N/m is the spring constant and x = 12cm = 0.12m is the initial displacement of the spring and A is the maximum displacement of the spring.

(b) when the speed is maximum, the kinetic energy is maximum, so the potential energy must be zero, then:

where v is the maximum speed.

(c) the energy dissipated per cycle is 0.03 J which is equal to the work done by the system W. The average power is defined as the rate of work.

So the average power P per cycle is :

P = W/T

where T is the time period of oscillation given by:

So,

P = 0.11 W

Explanation: