A string of a bow is pulled backwards 0.6 m with a force of 170 N. The mass of the arrow is 95 g (0.095 kg). If the bow is 92% efficient, at what speed will the arrow leave the bow?
I got 44 m/s
To solve this problem, we can use the principle of work and energy.
First, we need to find the potential energy stored in the bow when it is pulled backward. The potential energy (PE) is given by the formula:
PE = 0.5kx^2
Where k is the spring constant of the bow, and x is the distance the bowstring is pulled backward. Since we are given the distance (0.6 m) and the force (170 N), we can calculate the spring constant using Hooke's Law:
k = F / x
k = 170 N / 0.6 m
k = 283.33 N/m
Next, we can calculate the potential energy:
PE = 0.5 * 283.33 N/m * (0.6 m)^2
PE = 60.60 J (Joules)
Since the bow is 92% efficient, only 92% of the potential energy will be transferred to the arrow. Therefore, the arrow will have:
KE = 0.92 * 60.60 J
KE = 55.87 J
Finally, we can find the kinetic energy (KE) of the arrow using the formula:
KE = 0.5mv^2
Where m is the mass of the arrow and v is its velocity. Rearranging the formula, we can solve for v:
v = sqrt(2 * KE / m)
v = sqrt(2 * 55.87 J / 0.095 kg)
v = sqrt(1169.26 m^2/s^2 / 0.095 kg)
v = sqrt(12287.79 m^2/s^2)
v = 111 m/s (rounded to the nearest whole number)
Therefore, the speed at which the arrow will leave the bow is approximately 111 m/s, not 44 m/s.