Physics Question?

Question

A 0.49 kg object connected to a light spring with a spring constant of 22.3 N/m osciallates on a frictionless horizontal surface. The spring is compressed 7.1 cm and released from rest.

1) What is the maximum speed of the mass in m/s?

2) What is the speed (in m/s) of the object when the spring is compressed 1.4 cm?

3) What is the speed (in m/s) when the spring is stretched 1.4 cm?

4) For what value of x does the speed equal 1/2 of the maximum speed (in units of cm)?

Answer 1

This problem requires you to consider conservation of energy. When the spring is compressed the energy of the system is all potential. The potential energy is expressed as

PE = 0.5*k*x^2

where k is the spring constant and x is the compressed distance. When the spring is released the potential energy is converted to kinetic energy (0.5*m*v^2).

(1) The maximum speed is achieved when all the potential energy is converted to kinetic energy

0.5*k*xm^2 = 0.5*m*vm^2

Solving for the maximum speed (vm)

vm = sqrt(k/m)*xm

whenre k is the spring constant, m is the mass of the object, and xm is the maximum compression.

(2) When the object is at a position that is between maximum compression and maximum speed, the energy is a mixture of potential and kinetic.

0.5*k*xm^2 = 0.5*k*x^2 + 0.5*m*v^2

where x is the amount by which the spring is compressed and v is the speed at that point. Solving for v

v = sqrt((k/m)*(xm^2 - x^2)).

(3) The same rule applies when the spring is stretched rather than compressed.

0.5*k*xm^2 = 0.5*k*x^2 + 0.5*m*v^2

and

v = sqrt((k/m)*(xm^2 - x^2)).

Similarly, you could consider

0.5*m*vm^2 = 0.5*k*x^2 + 0.5*m*v^2

and solve for v

v = sqrt(vm^2 - (k/m)*x^2).

These will give you the same result.

(4) To find the position when the speed is 0.5*vm you apply the same logic.

0.5*m*vm^2 = 0.5*k*x^2 + 0.5*m*v^2

setting v = 0.5*vm and simplifying

m*vm^2 = k*x^2 + 0.25*m*vm^2

Solving for x

k*x^2 = m*(vm^2 - 0.25*vm^2) = 0.75*m*vm^2

x = sqrt(0.75*m/k)*vm