A luge and its rider, with a total mass of 110 kg, emerge from a downhill track

Question

A luge and its rider, with a total mass of 110 kg, emerge from a downhill track onto ahorizontal straight track with an initial speedof 70 m/s. Assume that they stopwith a constant deceleration of 1.9 m/s2. (a) What magnitude F is required for thedecelerating force?
N

(b) What distance d do they travel whiledecelerating?
m

(c) What work W is done on them by thedecelerating force?
kJ

. (d) What is F for a decelerationof 3.8 m/s2?
N

(e) What is d for a decelerationof 3.8 m/s2?
m

(f) What is W for a decelerationof 3.8 m/s2?
kJ
(a) What magnitude F is required for thedecelerating force?
N

(b) What distance d do they travel whiledecelerating?
m

(c) What work W is done on them by thedecelerating force?
kJ

. (d) What is F for a decelerationof 3.8 m/s2?
N

(e) What is d for a decelerationof 3.8 m/s2?
m

(f) What is W for a decelerationof 3.8 m/s2?
kJ

Explanation / Answer

a) since a = -1.9m/s2 the force F = ma = 110kg*(-1.9m/s2) =-209N   negative means is in opposite direction ofmotion b) since vf = 0 when it stop, vi = 70m/s,let distance be d, we have: vf2 - vi2 =2ad d = (vf2 - vi2)/2a= (0-(70m/s)2)/(2*(-1.9m/s2)) =1289.47m c) W = Fd = -209N*1289.47 = -269.5 kJ d) F=ma=110kg*(-3.8m/s2) = -418N e) d = (vf2 - vi2)/2a= (0-(70m/s)2)/(2*(-3.8m/s2)) = 644.74m f) W = Fd = -418N*644.74 = -269.5 kJ

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---