Spring equilibrium

1. Nov 27, 2005

cgotu2

A 1.5 kg block is attached to a spring with a spring constant of 1950 N/m. The spring is then stretched a distance of 0.33 cm and the block is released from rest.
(a) Calculate the speed of the block as it passes through the equilibrium position if no friction is present.

(b) Calculate the speed of the block as it passes through the equilibrium position if a constant frictional force of 2.0 N retards its motion.

(c) What would be the strength of the frictional force if the block reached the equilibrium position the first time with zero velocity?

I got part (a) by by saying that 1/2mv^2=1/2kA^2 and my answer of 11.898 cm/s was correct. but i can't get (b) or (c). in (b) i said that the work done by friction (2*0.33) was equal to 1/2mv^2 - 1/2kA^2, but my answer is incorrect.

2. Nov 27, 2005

andrewchang

for part b, the work done by friction is negative

3. Nov 27, 2005

cgotu2

okay i did that, and i got 11.86 cm/s, but it is still wrong (for b)

4. Nov 27, 2005

andrewchang

it's stretched 0.33 centimeters?

make sure you use the right units, i guess..

5. Nov 28, 2005

andrevdh

From the work-kinetic energy theorem we have that the work done by all of the forces on a system is equal to the change in i'ts kinetic energy or
$$W=\Delta K$$
We can split the work done into the two basis types for conservative an non conservative forces
$$W_{nc} \ + \ W_c=\Delta K$$
and since the work done by the conservative force is equal to the negative of the change in the sytem's potential energy or
$$W_c=-\Delta U$$
we can change the previous relation to
$$W_{nc} = \Delta K \ + \ \Delta U$$
In this case the non conservative force is friction.

The last question is actually quite easy if one can answer this question: How much energy must be removed from the system, by the frictional force, in order to bring the mass to a halt at the equilibrium position?

Last edited: Nov 28, 2005