A. 9 J
In a force-distance graph, the work done is equal to the area under the curve in the graph.
In this case, we need to extrapolate the value of the force when the distance is x=30 cm. We can easily do that by noticing that there is a direct proportionality between the force and the distance:
[tex]F=kx[/tex]
where k is the slope of the line. We can find k, for instance chosing the point at x=5 cm and F=10 N:
[tex]k=\frac{F}{x}=\frac{10 N}{5 cm}=2 N/cm[/tex]
And now we can calculate the work by calculating the area under the curve until x=30 cm, F=60 N:
[tex]W=\frac{1}{2} (height) (base)= \frac{1}{2}(60 N)(0.30 m)=9 J[/tex]
B. 24.5 m/s
The mass of the arrow is m=30 g=0.03 kg. The kinetic energy of the arrow when it is released is equal to the work done by pulling back the bow for 30 cm:
[tex]W=K=\frac{1}{2}mv^2[/tex]
where m is the mass of the arrow and v is its speed. By re-arranging the formula and using W=9 J, we find the speed:
[tex]v=\sqrt{\frac{2W}{m}}=\sqrt{\frac{2\cdot 9J}{0.03 kg}}=24.5 m/s[/tex]
