Answer:
a= -5404.6 [m/s²]; F=785.75 [N].
Explanation:
1. Determine is the acceleration of the ball:
equation of the distance (0.135m) is:
[tex]S=-\frac{at^2}{2} +v_0t,[/tex] Â where S=0.135[m]; a - required acceleration; t - elapsed time; vâ‚€ - initial velocity (38.2 m/s);
also the required acceleration is:
[tex]a=\frac{v-v_0}{t}, where[/tex] V - the end velocity (0 m/s), t - elapsed time, â‚€ - initial velocity (38.2 m/s).
Using the equations of reqruired acceleration and the distance it is possible to make up and solve the next system:
[tex]\left \{ {{a=-\frac{38.2}{t} } \atop {S=-\frac{at^2}{2}+38.2t }} \right. \ => \ \left \{ {{a=-\frac{38.2}{t} } \atop {0.5at^2-38.2t=0.135}} \right. \ => \ \left \{ {{a=-\frac{38.2}{t} } \atop {19.1t-38.2t=0.135}} \right. \ => \ \left \{ {{a=-5404.6} \atop {t=0.007}} \right.[/tex]
finally, a≈-5404.6 [m/s²].
2. Determine the force applied by Alejandro.
the energy is:
[tex]E=\frac{mv^2}{2}; \ or E=FS, where \ m-the \ mass; \ v-velocity; \ F-required \ force; S-distance;[/tex]
According to these two equations, the required force is:
[tex]F=\frac{E}{S}=\frac{mv^2}{2S};[/tex]
F=0.145*38.8²/2/0.135≈785.75 [N].
note, the suggested way is not the shortest one and not the only one.
P.S. if it is possible, check the arithmetic operations and the provided answers in other sources.
