In preparation for shooting a ball in a pinball machine,a spring (K=675N/m) is compressed by 0.0690m relative to its unstrained length. The ball(m=0.600kg) is at rest against the spring at point A.When the spring is released, the ball slides (without rolling) to point B, which is 0.300m higher than point A. How fast is the ball moving at B?
solution:Ef=E0
1/2 MVsq+1/2kXfsq =1/2kx0sq
vf= sqrootk/m(Xintial sq+xfinal sq)
vf= [675/0.600kg(0.690m+.300m)]sqroot
vf=20.37m/s2
I don't think it is right
No. Your algebra is in error (sign).
I get something like..
v2=k/m (xf2 - xo2 )
then solve for v.
Ah, algebra errors can be quite sneaky, can't they? Let's take a closer look at your solution.
First, let's correct the equation: v^2 = (k/m) * (xf^2 - xo^2)
Now, let's substitute the given values into the equation:
v^2 = (675 N/m / 0.600 kg) * ((0.0690 m + 0.300 m)^2 - 0^2)
Simplifying:
v^2 = (675 N/m / 0.600 kg) * (0.3690 m)^2
v^2 = (675 N/m / 0.600 kg) * (0.136161 m^2)
v^2 = 1501.5 m^2/s^2
Taking the square root of both sides:
v = √(1501.5) m/s
v ≈ 38.74 m/s
So, the ball is moving at approximately 38.74 m/s when it reaches point B.
To solve for the speed of the ball at point B, we can use the principle of conservation of energy. The initial potential energy stored in the compressed spring is equal to the kinetic energy of the ball at point B.
The equation for conservation of energy is:
E_f = E_i
The initial potential energy in the compressed spring is given by:
E_i = (1/2)k(x_i)^2
where k is the spring constant (675 N/m) and x_i is the initial compression of the spring (0.0690 m).
The final kinetic energy of the ball at point B is given by:
E_f = (1/2)mv^2
where m is the mass of the ball (0.600 kg) and v is the speed of the ball at point B (what we are trying to solve for).
Setting these two equations equal to each other:
(1/2)k(x_i)^2 = (1/2)mv^2
Simplifying and rearranging for v:
v^2 = k/m * (x_i)^2
Plugging in the given values:
v^2 = (675 N/m) / (0.600 kg) * (0.300 m + 0.0690 m)^2
Calculating:
v^2 = 675 N/m / 0.600 kg * (0.369 m)^2
v^2 = 675 N/m / 0.600 kg * 0.136161 m^2
v^2 = 1490.38 m^2/s^2
Finally, taking the square root to solve for v:
v = sqrt(1490.38 m^2/s^2)
v ≈ 38.62 m/s
Therefore, the ball is moving at a speed of approximately 38.62 m/s at point B.
To solve this problem, we can use the principle of conservation of mechanical energy.
The initial potential energy stored in the compressed spring is equal to the final kinetic energy of the ball at point B.
First, let's calculate the potential energy stored in the spring when it is compressed by 0.0690m.
The potential energy stored in a spring can be calculated using the formula:
Potential energy = (1/2)kx^2
where k is the spring constant (675 N/m) and x is the displacement from the equilibrium position (0.0690m).
Potential energy = (1/2)(675 N/m)(0.0690m)^2
Potential energy = 16.53 J
Next, let's calculate the change in the potential energy of the ball as it moves from point A to point B.
The change in potential energy, ΔPE, is equal to the mass of the ball (0.600 kg) multiplied by the acceleration due to gravity (9.8 m/s^2) and the change in height (0.300 m).
ΔPE = mgh
ΔPE = (0.600 kg)(9.8 m/s^2)(0.300 m)
ΔPE = 1.764 J
Now, we can equate the initial potential energy to the final kinetic energy of the ball:
Potential energy at A = Kinetic energy at B
Now, let's solve for the final velocity of the ball, v.
(1/2)mv^2 = ΔPE
(1/2)(0.600 kg)v^2 = 1.764 J
v^2 = (2 * 1.764 J) / 0.600 kg
v^2 = 5.88 J/kg
v = √(5.88 J/kg)
v ≈ 2.42 m/s
Therefore, the ball is moving at approximately 2.42 m/s at point B.