The 87.8-kg biker shown picks up speed as he rides down a semicircular incline f
ID: 1849418 • Letter: T
Question
The 87.8-kg biker shown picks up speed as he rides down a semicircular incline from position A leading up to a jump at B. The radius of curvature of the slope R is 31 m and the biker is angled at theta1 = 25.6 degree s to the horizontal at the instant he jumps. The effects of drag and of friction in the bike's axle and gears are negligible, and the biker does not pedal or brake. On his first approach, the biker reaches a velocity of 16.3 m/s as he approaches B. Determine the magnitude of the normal force exerted by the slope on the biker just before he passes through B. (N) This time the biker starts from rest at position A, located theta2 = 39.1 degree s up the slope from its lowest point at C. As he rides down the slope and up to the jump he travels against the wind, which exerts a constant 7.8 N force on him horizontally to the left. Calculate the normal force exerted by the slope on the biker just before he passes B. (N) Incorrect Marks for this submission: 0/3.Explanation / Answer
a) Normal force = (mVb^2/R) + mgCos25.6 = 87.8*16.3^2/31 + 87.8*9.81*Cos25.6 = 1529.267 N
b) Potential energy change from A to B = mg (R - R Cos39.1) - mg (R - RCos25.6) = mgR (Cos 25.6 - Cos39.1)
Horizontal distance from A to B = R Sin39.1 + R Sin25.6 = R (Sin39.1 + Sin25.6)
Energy lost against wind = 7.8*R (Sin39.1 + Sin25.6)
K.E at B = mgR (Cos 25.6 - Cos39.1) - 7.8*R (Sin39.1 + Sin25.6)
1/2*m*Vb^2 = mgR (Cos 25.6 - Cos39.1) - 7.8*R (Sin39.1 + Sin25.6)
1/2*m*Vb^2 = 87.8*9.81*31 (Cos 25.6 - Cos39.1) - 7.8*31* (Sin39.1 + Sin25.6)
1/2*m*Vb^2 = 3358.6 - 257 = 3101.6 J
mVb^2 = 2*3101.6 = 6203.2 J
Normal force = (mVb^2/R) + mgCos25.6 = 6203.2/31 + 87.8*9.81*Cos25.6 = 976.869 N
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.