Answer:
1.40 m/s
Explanation:
The potential energy of a compressed spring can be expressed as:
[tex]E_{ps}=\dfrac{1}{2}kx^2[/tex]
From above;
k = spring constant
x = distance of the spring (compressed)
From the barrel, the kinetic energy (i.e. the final K.E) of the ball is calculated using the relation:
[tex]E_{kf}= \dfrac{1}{2}mv^2[/tex]
where;
m = the ball mass
v = ball's speed
Equating both equations above, we have:
[tex]E_{ps}- F_fd=E_{kf[/tex]
This can be re-written as:
[tex]\dfrac{1}{2}kx^2 - F_fd=\dfrac{1}{2}mv^2}[/tex]
[tex]v^2 = (\dfrac{k}{m})x^2-\dfrac{2F_fd}{m}[/tex]
[tex]v =\sqrt{ (\dfrac{k}{m})x^2-\dfrac{2F_fd}{m}}[/tex]
replacing the values from the given information:
[tex]v =\sqrt{ (\dfrac{8.00\ N/m}{5.30\times10^{-3} \ kg})(5.00 \ cm \times \dfrac{10^{-2} \ m}{1 \ cm})^2-(\dfrac{2(0.032 \ N)(0.150 \ m)}{5.30\times \dfrac{10^{-3} \ kg}{1 \ g}})}[/tex]
[tex]v = \sqrt{1.962264151}[/tex]
v ≅ 1.40 m/s
The speed at which the projectile leaves the barrel of the cannon will be given as [tex]v=1.40\ \frac{m}{s}[/tex]
Speed is defined as the movement of any object with respect to time. It is the ratio of distance and time.
Now it is given in the question:
Mass of ball m = 5.30 g
The deflection of spring = 5 cm
The force constant of spring [tex]k= 8 \ \frac{N}{m^2}[/tex]
The length of the barrel = is 15 cm
The frictional force of the barrel = 0.032 N
Now from the conservation of energy, we can write as
[tex]E_{spring}-E_{friction}=E_{ball}[/tex]
[tex]\dfrac{1}{2} kx^2-F_fd=\dfrac{1}{2} mv^2[/tex]
[tex]v=\sqrt{\dfrac{k}{m}(x^2) -\dfrac{2F_fd}{m} }[/tex]
Now putting the values in the above formula:
[tex]v=\sqrt{\dfrac{8}{5.30\times 10^{-3}}(15\times10^{-2}) -\dfrac{2\times(0.0032)\times (0.015)}{5.30\times 10^{-3}} }[/tex]
[tex]v=1.40\ \frac{m}{s}[/tex]
Thus the speed at which the projectile leaves the barrel of the cannon will be given as [tex]v=1.40\ \frac{m}{s}[/tex]
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