Question
An archer puts a 0.28 kg arrow to the bow string. An average force of 200 N is exerted to draw the sting back 1.1 m. a) Assuming no frictional loss, what speed does the arrow leave the bow? b) If the arrow is shot straight up, how high does it rise?
My thoughts: KE = 1/2mv^2. You are given mass but how do you find velocity? Once velocity is given KE = PE = mgh.
My thoughts: KE = 1/2mv^2. You are given mass but how do you find velocity? Once velocity is given KE = PE = mgh.
Answers
(a) Work done drawing the arrow back
= 200 N x 1.1 m
=(1/2) M V^2
Solve for V
(b) Knowing V, the maximum height attained (h) is given by
(V^2/2g) = h
= 200 N x 1.1 m
=(1/2) M V^2
Solve for V
(b) Knowing V, the maximum height attained (h) is given by
(V^2/2g) = h