A .20 kg ball is attached to a vertical spring. The spring constant is 28 N/m. The ball is supported initially so that the spring is neither stretched or compressed, is released from rest. How far does the ball fall before brought to a momentary stop by the spring?
Our teacher gave us a hint:
PE+KE=PE+KE
But I think that there is something missing, perhaps subscripts (initial and final). And if that's the case, then, I'll get PE=PE.
Also, someone else in class explained that PE=mgh and solved the problem, but I do not know how she got .5 for h.
Please help, so confused. Thank you.
ETi = ETf
Ek+ Eg+Ee = Ek+Eg+Ee
since vi and vf is o Ek is taken out, and having a ref point at the bottom Eg is for ETf is taken out( final hight =0 meters). Also spring is at equilibirum in the beinging, so Ee for ETi is also taken out( xi = O meters)
so then Eg= Ee
and mghi= 0.5(k)(xf)^2
ur hi= xf
mghi= 0.5(k)(hi)^2
hi=mg/0.5(k)
hi= (0.20)(9.80)/(0.5)(28)
hi= 0.14
wats weird is my gr.12 physics teacher gave the same exact question, kk hope it helped! good luck
To solve this problem, we need to apply the conservation of mechanical energy principle, which states that the total mechanical energy of a system remains constant if no external forces are acting on it. In this case, we can consider the system as the ball and the spring.
Let's break down the problem step by step:
Step 1: Initial condition
When the ball is supported and not stretched or compressed, its total mechanical energy is initially zero because both potential energy (PE) and kinetic energy (KE) are zero.
Step 2: Falling motion
As the ball falls, it gains potential energy due to its height above the ground. The potential energy can be calculated using the formula: PE = mgh, where m is the mass of the ball, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height.
Step 3: Momentary stop by the spring
When the ball is brought to a momentary stop by the spring, it has converted all of its potential energy into potential energy stored in the spring. The potential energy stored in a spring is given by the equation: PE = (1/2)kx^2, where k is the spring constant and x is the displacement of the spring from its equilibrium position.
Step 4: Equating energies
According to the conservation of mechanical energy, the initial mechanical energy is equal to the final mechanical energy, expressed as PE_initial + KE_initial = PE_final + KE_final.
Since the ball is initially at rest, its initial kinetic energy is zero. Hence, the equation becomes: 0 + 0 = PE_final + KE_final.
Substituting the formulas for the potential energies, the equation becomes: 0 + mgh = (1/2)kx^2 + 0.
However, to solve for x, we need to find the value of h in terms of x, which represents the distance the ball falls before being stopped by the spring.
In this case, the ball would be momentarily at rest when the spring is fully compressed. The distance the spring is compressed would be the same as the distance the ball falls from its initial position. Let's call this distance x.
So, we can rewrite the equation as: mgh = (1/2)kx^2.
Now, let's substitute the given values into the equation and solve for x:
m = 0.20 kg
g = 9.8 m/s^2
h = x (assuming the height fallen is equal to the distance the spring is compressed)
k = 28 N/m
0.20 kg * 9.8 m/s^2 * x = (1/2) * 28 N/m * x^2
Simplifying the equation:
1.96 x = 14x^2
Rearranging the equation:
14x^2 - 1.96x = 0
Now we can solve this quadratic equation for x using factoring or the quadratic formula:
14x(x - 0.14) = 0
This equation has two possible values for x: x = 0 and x = 0.14 m.
Since x = 0 represents the uncompressed position of the spring, the distance the ball falls before being brought to a momentary stop by the spring is x = 0.14 m.
Therefore, the ball falls 0.14 meters before being stopped by the spring.