An arrow with a mass of 0.0180 kg enters a block of foam traveling at 138 ms and comes to a stop in 0.0042 s. The arrow entered into the block at a distance of 0.2898 m. What is the acceleration?

Respuesta :

Answer:

[tex]a = 32857.167\,\frac{m}{s^{2}}[/tex]

Explanation:

The motion of the arrow can be described by the Moment Conservation and Impact Theorem:

[tex](0.018\,kg)\cdot \left(138\,\frac{m}{s} \right) - F\cdot (0.0042\,s) = (0.018\,kg)\cdot \left(0\,\frac{m}{s} \right)[/tex]

The impact force is computed herein:

[tex]F = 591.429\,N[/tex]

The deceleration experimented by the arrow is:

[tex]a = \frac{F}{m}[/tex]

[tex]a = \frac{591.429\,N}{0.018\,kg}[/tex]

[tex]a = 32857.167\,\frac{m}{s^{2}}[/tex]