How Fast Will the Block Move After Being Released from the Spring?

[volraithe]

A 1.54-kg block is held against a spring of force constant 1.47E+4 N/m, compressing it a distance of 0.100 m. How fast is the block moving after it is released and the spring pushes it away?

i thought u just use the formulas that F=kx and then plug that into F=ma, but i am not getting the right answer... anyone have any other suggestions as to what path I should take?
thank you.

 

[Pengwuino]

A 1.54-kg block is held against a spring of force constant 1.47E+4 N/m, compressing it a distance of 0.100 m. How fast is the block moving after it is released and the spring pushes it away?

i thought u just use the formulas that F=kx and then plug that into F=ma, but i am not getting the right answer... anyone have any other suggestions as to what path I should take?
thank you.

Doing that will only give you the instantaneous acceleration. The problem with this is that the force will change continuously as the mass is pushed with the spring; this results in a constantly changing acceleration and we can't use our kinematics very easily. We want to use something else.

A very simple method for this problem would be to use the fact that the potential energy of the spring is $$\frac{{kx^2 }}{2}$$. We know that upon release, the spring is going to transfer all of it's energy into the mass. We of course know the energy of a mass as $$\frac{{mv^2 }}{2}$$. Equating both sides allows you to solve for the velocity of the mass.

 

[nlightner]

Based on the given information, we can use the formula F = kx to calculate the force exerted by the spring on the block. Plugging in the values, we get F = (1.47E+4 N/m)(0.100 m) = 1470 N.

Next, we can use the formula F = ma to calculate the acceleration of the block. Since the spring is the only force acting on the block, we can set F = ma and solve for a. Thus, a = F/m = 1470 N/1.54 kg = 954.55 m/s^2.

Using the kinematic equation v^2 = u^2 + 2as, where u is the initial velocity (0 m/s), we can solve for the final velocity of the block after it is released by setting v = ?, s = 0.100 m, and a = 954.55 m/s^2.

Thus, v^2 = 0^2 + 2(954.55 m/s^2)(0.100 m) = 190.91 m^2/s^2. Taking the square root of both sides, we get v = 13.81 m/s. Therefore, the block will be moving at a speed of 13.81 m/s after it is released by the spring.

If this answer does not match the expected result, double check your calculations and make sure you are using the correct units. It is also important to note that this calculation assumes no external forces acting on the block, and that the spring is ideal and does not have any internal damping.