## The spring of a spring gun has force constant k =400 N/m and negligible mass. The spring is compressed6.00 cm and a ball with mass 0.0300 kg

Question

The spring of a spring gun has force constant k =400 N/m and negligible mass. The spring is compressed6.00 cm and a ball with mass 0.0300 kg isplaced in the horizontal barrel against the compressed spring. Thespring is then released, and the ball is propelled out the barrelof the gun. The barrel is 6.00 cm long, so the ballleaves the barrel at the same point that it loses contact with thespring. The gun is held so the barrel is horizontal.
Calculate the speed with which the ballleaves the barrel if you can ignore friction.
Calculate the speed of the ball as it leavesthe barrel if a constant resisting force of 6.00 Nacts on the ball as it moves along the barrel.
For the situation in part (b), at whatposition along the barrel does the ball have the greatest speed?(In this case, the maximum speed does not occur at the end of thebarrel.)
What is that greatest speed?

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1 year 2021-09-02T21:48:41+00:00 1 Answers 13 views 0

1. a) 6.9 m/s

b) 4.9 m/s

c) After 4.50 cm

d) 5.2 m/s

Explanation:

a)

In this case, there is no resistive force. Therefore, according to the law of conservation of energy, all the initial elastic potential energy stored in the spring when it is compressed is converted into kinetic energy of the ball as it leaves the barrel; so we can write:

$$\frac{1}{2}kx^2=\frac{1}{2}mv^2$$

where:

k = 400 N/m is the spring constant

x = 6.00 cm = 0.06 m is the compression of the spring

m = 0.03 kg is the mass of the ball

v is the velocity of the ball as it leaves the barrel

Solving the equation, we can find the speed of the ball:

$$v=\sqrt{\frac{kx^2}{m}}=\sqrt{\frac{(400)(0.06)^2}{0.03}}=6.9 m/s$$

b)

In this case, there is a constant resistive force of

$$F_r = 6.00 N$$

acting on the ball as it moves along the barrel.

The work done by this force on the ball is:

$$W=-F_rd$$

where

d = 6.00 cm = 0.06 m is the length of the barrel

And where the negative sign is due to the fact that the force is opposite to the motion of the ball

Substituting,

$$W=-(6.00)(0.06)=-0.36 J$$

Therefore, part of the initial elastic potential energy stored in the spring has been converted into thermal energy due to the resistive force. So, the final kinetic energy of the ball will be less than before:

$$\frac{1}{2}kx^2+W=\frac{1}{2}mv^2$$

And solving for v, we find the new speed:

$$v=\sqrt{\frac{kx^2}{m}+\frac{2W}{m}}=4.9 m/s$$

c)

The (forward) force exerted by the spring on the ball is

$$F=k(x_0-x)$$

where

k = 400 N/m

x is the distance covered by the ball

$$x_0=0.06 m$$ is the maximum displacement of the spring

While the (backward) resistive force is instead

$$F_r=6.0 N$$

So the net force on the ball is

$$F=k(x-x_0)-F_r$$

The term $$k(x-x_0)$$ prevails at the beginning, so the ball continues accelerating forward, until this term becomes as small as $$F_r$$: after that point, the negative resistive force will prevail, so the ball will start delecerating. Therefore, the greatest speed is reached when the net force is zero; so:

$$k(x_0-x)-F_r=0\\x=x_0-\frac{F_r}{k}=0.06-\frac{6.00}{400}=0.045 m$$

d)

After the ball has covered a distance of

x = 0.045 m

The work done by the resistive force so far is:

$$W=-F_r x =-(6.00)(0.045)=-0.27 J$$

The total elastic potential energy of the spring at the beginning was

$$E_e=\frac{1}{2}kx_0^2 = \frac{1}{2}(400)(0.06)^2=0.72 J$$

While the elastic potential energy left now is

$$E_e’=\frac{1}{2}k(x_0-x)^2=\frac{1}{2}(400)(0.015)^2=0.045 J$$

So, the kinetic energy now is:

$$K=E_e-E_e’+W=0.72-0.045-0.27=0.405 J$$

And by using the equation

$$K=\frac{1}{2}mv^2$$

We can find the greatest speed:

$$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(0.405)}{0.03}}=5.2 m/s$$