Spring / constant applied force question

[kennien]

Hi

A block ( 0.5 kg ) is attached to a spring ( k = 40N/m ) on a frictionless table..

The spring is 0.6 meter long unstretched ..

A constant force (20 N ) is applied horisontaly causing the spring to stretch

1) question : what is the speed/velocity of the box after 0.25 meter ?

can anyone help me with this ?

I understand that the box accelerates according to F=ma in the beginning , but when the spring starts to exert force in the other direction , that's when I am loosing it

 

[physics girl phd]

draw a free body diagram of the box when both your external force and the spring are acting on it. The diagram stays the same, even though the force of the spring changes as a function of position. Use this diagram to find F as a function of x. I'd personally approach this as a work/energy problem... using the expression for force and integrating that expression over the path of motion. then you know the energy into the system. Any energy NOT converted to potential energy in the spring remains as kinetic energy, which you can use to find speed.

 

[baba89]

.

Sure, I'd be happy to help you with this question! First, let's review the basics of springs and forces. A spring is a type of elastic material that can be stretched or compressed when a force is applied to it. The constant, k, represents the stiffness of the spring and how much force it takes to stretch the spring a certain distance. In this case, the spring has a stiffness of 40N/m, meaning that it takes 40N of force to stretch the spring by 1 meter.

Now, let's look at the situation described in the question. A 0.5 kg block is attached to the spring and a constant force of 20N is applied horizontally. This force will cause the spring to stretch and the block to accelerate. We can use Newton's Second Law, F=ma, to calculate the acceleration of the block. Since we know the mass (0.5 kg) and the force (20N), we can rearrange the equation to solve for acceleration: a = F/m = 20N/0.5kg = 40m/s^2.

Next, we need to consider the distance that the block has traveled after 0.25 meters. We know that the spring is 0.6 meters long when unstretched, so after 0.25 meters, the spring will have stretched to a length of 0.85 meters (0.6m + 0.25m). At this point, the spring will start to exert a force in the opposite direction of the applied force, due to its elasticity. This means that the acceleration of the block will start to decrease.

To calculate the speed of the block after 0.25 meters, we can use the equation v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity (which is 0 since the block starts from rest), a is the acceleration, and s is the distance traveled. Plugging in the known values, we get v^2 = 0 + 2(40m/s^2)(0.25m) = 20m^2/s^2. Taking the square root, we get v = 4.47 m/s.

I hope this helps to clarify the situation for you. Remember, when dealing with springs and forces, it's important to consider the distance traveled and the elastic properties of the spring. Let me know if you