Respuesta :
a) The speed of the ball as it leaves the cannon is 1.44 m/s
b) The ball has maximum speed at 5.08 cm from the initial position (equilibrium position of the spring)
c) The maximum speed of the ball is 1.83 m/s
Explanation:
a)
We can solve this part of the problem by using the law of conservation of energy.
In fact, the total initial energy of the system is equal to the elastic potential energy stored in the spring when it is fully compressed, therefore:
[tex]E=\frac{1}{2}kx^2[/tex]
where
k = 8.07 N/m is the spring constant
x = 5.08 cm = 0.0508 m is the compression of the spring
Substituting,
[tex]E=\frac{1}{2}(8.07)(0.0508)^2=0.0104 J[/tex]
When the ball is fired, part of this energy is converted into kinetic energy of the ball, while the rest is converted into thermal energy due to the work done by the frictional force:
[tex]E=K+W[/tex] (1)
where:
K is the kinetic energy of the ball
W is the work done by friction
The work done by friction is
[tex]W=Fd=(0.0313)(0.158)=0.0049 J[/tex]
where F = 0.0313 N is the force of friction and d = 15.8 cm = 0.158 m is the distance covered by the ball while in the cannon
Using (1), we now find the kinetic energy of the ball as it leaves the cannon:
[tex]K=E-W=0.0104 - 0.0049 = 0.0055 J[/tex]
The kinetic energy can be rewritten as
[tex]K=\frac{1}{2}mv^2[/tex]
where
m = 5.28 g = 0.00528 kg is the mass of the ball
v is its speed as it leaves the cannon
Solving for v,
[tex]v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(0.0055)}{0.00528}}=1.44 m/s[/tex]
b)
As we stated before, the total mechanical energy of the system (the elastic potential energy initially stored in the spring) is then converted into kinetic energy and thermal energy:
[tex]E=K+W[/tex]
This means that the kinetic energy (and so, the speed) is maximum when all the initial elastic potential energy has been released.
The elastic potential energy is given by
[tex]E=\frac{1}{2}kx^2[/tex]
where
k is the spring constant
x is the displacement
The elastic potential energy is all released when the displacement becomes zero, x = 0: therefore, the ball has maximum kinetic energy and maximum speed when it passes through the equilibrium position of the spring (so, at 5.08 cm from the initial position).
c)
For this part we can use again
[tex]E=K+W[/tex]
where:
E = 0.0104 J is the total mechanical energy
K is the kinetic energy at the equilibrium position
And the work done by friction in the first 5.08 cm of motion is:
[tex]W=Fd'=(0.0313 N)(0.0508 m)=0.0016 J[/tex]
So the kinetic energy when the ball transits the equilibrium position is
[tex]K=E-W=0.0104-0.0016=0.0088 J[/tex]
And so the maximum speed is
[tex]v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(0.0088)}{0.00528}}=1.83 m/s[/tex]
Learn more about kinetic energy here:
brainly.com/question/6536722
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