Question

Compressed air is used to fire a 60 g ball vertically upward from a 0.70-m-tall tube. The air exerts an upward force of 3.0 N on the ball as long as it is in the tube. Part A How high does the ball go above the top of the tube

Answer 1

Answer:

2.87 m

Explanation:

Given:

Mass of the ball (m) = 60 g = 0.06 kg

Height of the tube (h) = 0.70 m

Upward force by compressed air on the ball (F) = 3.0 N

Initial velocity of the ball (u) = 0 m/s (Assume)

Final velocity of the ball at the end of tube (v)  = ?

Acceleration of the ball (a) = ?

Weight of the ball is equal to the product of mass and gravity. So,

Weight (W) = $mg=0.06\times 9.8=0.588\ N$

Therefore, the net force acting on the ball is equal to the difference of upward force and downward force. So,

Net force = Force by air - Weight

$F_{net}=F-mg\\F_{net}=3.0-0.588 = 2.412\ N$

Now, from Newton's second law, the net force is equal to the product of object's mass and acceleration. So,

$F_{net}=ma\\\\a=\frac{F_{net}}{m}=\frac{2.412\ N}{0.06\ kg}=40.2\ m/s^2$

Now, acceleration (a) = 40.2 m/s²

Using equation of motion, we have:

$v^2=u^2+2ah\\\\v^2=0+2\times 40.2\times 0.7\\\\v=\sqrt{56.28}=7.5\ m/s$

Now, let the maximum height reached be 'H'.

So, applying energy conservation from the top of pipe to the maximum height.

Decrease in kinetic energy = Increase in potential energy.

$\frac{1}{2}mv^2=mgH\\\\H=\frac{v^2}{2g}$

Plug in the given values and solve for 'H'. This gives,

$H=\frac{56.28}{2\times 9.8}\\\\H=\frac{56.28}{19.6}=2.87\ m$

Therefore, the ball reaches a height of 2.87 m above the top of the tube.