The graph at the right shows the force needed to pull a bow back as the string is pulled further and further. a. how much work is done pulling the bow back to 30cm? b. if a 30g arrow is shot from the bow, how fast will the arrow be going when it leaves the bow?

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:

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:

And now we can calculate the work by calculating the area under the curve until x=30 cm, F=60 N:

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:

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:

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft