An archer shoots a 100 gram arrow straight up in the air. thebow was drawn back
ID: 1721633 • Letter: A
Question
An archer shoots a 100 gram arrow straight up in the air. thebow was drawn back 50 cm by a 150 N force. how high does the arrowrise assuming friction was neglibible? How high does the arrow riseassuming friction is negligible? How high does the arrow riseassuming the force of air friction on the arrow averages 0.080during teh upward flight. assume the bow acts like an idealspring. I get -(150N/.50 m)=300 N/m = K v2=84.8 m/s v=9.20m/s I am lost...help! An archer shoots a 100 gram arrow straight up in the air. thebow was drawn back 50 cm by a 150 N force. how high does the arrowrise assuming friction was neglibible? How high does the arrow riseassuming friction is negligible? How high does the arrow riseassuming the force of air friction on the arrow averages 0.080during teh upward flight. assume the bow acts like an idealspring. I get -(150N/.50 m)=300 N/m = K v2=84.8 m/s v=9.20m/s I am lost...help!Explanation / Answer
a. Springconstant k = force/ extension in string = 150 /0.50 = 300 N/m Potentialenergy stored in bow U = (1/2)* k * x2 = 0.5* 300 *0.502 = 37.50 J According to law of conservation of energy,this must be equal to potential energy of arrow at highest pointi.e. U = m* g * h h = 37.50/ 0.100 * 9.8 = 38.26 m b. Retardation produced inarrow due to friction ofair aair = Ffriction/ m = 0.080 /0.100 = 0.80 m/s2 Hence netdeacceleration a = g + aair = 9.8+0.80 = 10.60 m/s2 therefore U = m* a * H Maximum height reached with airfriction H = 37.50/ 0.100 * 10.60 = 35.38 m therefore U = m* a * H Maximum height reached with airfriction H = 37.50/ 0.100 * 10.60 = 35.38 mRelated Questions
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