1) a runner moving at his speed of 4.5 m/s rounds a curved track with a 64 m radius. what is the runner's centripetal acceleration?
a. 0.07 m/s^2
b. 0.316 m/s^2
c. 288 m/s^2
d. 1296 m/s^2
2) a 58.0 kg skater with a velocity of 16.0 m/s comes into a curve of 24.0 m radius. how much friction must be exerted between the skates and the ice to negotiate the curve?
a. 2088 N
b. 38.7 N
c. 87 N
d. 618.7 N
2007-11-29 09:26:32 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Physics
1. a = V^2/r = 4.5^2/ 64 = .316 m/s^2
2. mv^2/r = 58(16^2/24) = 618 N
2007-11-29 09:54:39 · answer #1 · answered by Helper 6 · 0⤊ 0⤋
(a) Centripetal acceleration can be computed as a = v^2 / r. You have v and r; compute a.
(b) The friction from the skates must provide the centripetal force that keeps the skater on the curve. Compute a as in part (a) above, then find the force from Newton's Second Law F = ma.
2007-11-29 17:31:49 · answer #2 · answered by jgoulden 7 · 0⤊ 0⤋