Question

The skiing duo of Brian (77 kg ) and Ashley (54 kg ) is always a crowd pleaser. In one routine, Brian, wearing wood skis, starts at the top of a 230-mm-long, 20∘∘ slope. Ashley waits for him halfway down. As he skis past, she leaps into his arms and he carries her the rest of the way down. Hint: Use coefficient of kinetic friction from Table 6.1 in the textbook.

wood on snow kinetic friction is 0.06

What is their speed at the bottom of the slope?

Answer 1

mb = mass of brian = 77 kg

ma = mass of ashley = 54 kg

L = length of the slope = 230 mm = 0.230 m

= Angle of slope = 20o deg

vb = speed of brian at the location of ashley just before collision

va = speed of ashley just before collision = 0 m/s

v = combined speed after the collision

hb = height from which brian starts = L Sin = (0.230) Sin20

ha = height from which ashley leaps into arms of brian = (L/2) Sin

= coefficient of kinetic friction = 0.06

Using conservation of energy for the motion of brian from starting point to the location of ashley :

Initial potential energy at starting point + work done by kinetic frictional force = Final potential energy at ashley's location + Kinetic energy of brian

mb g hb + ( mb g Cos) d Cos180 = mb g ha + (0.5) mb vb2

(77) (9.8) (0.230) Sin20 + ((0.06) (77) (9.8) Cos20) (0.230/2) (-1) = (77) (9.8) (0.230/2) Sin20 + (0.5) (77) vb2

vb = 0.8 m/s

Using conservation of momentum

ma va + mb vb = (ma + mb) v

(54) (0) + (77) (0.8) = (54 + 77) v

v = 0.47 m/s

Using conservation of energy for the combination from the location of collision to the bottom of the incline

Initial potential energy just after collision + work done by kinetic frictional force + kinetic energy just after collision = Kinetic energy at bottom

(ma + mb) g (L/2) Sin + ( mb g Cos) d Cos180 + (0.5) (ma + mb) v2 = (0.5) (ma + mb) v'2

(77 + 54) (9.8) (0.230/2) Sin20 + ((0.06) (77 + 54) (9.8) Cos20) (0.230/2) (-1) + (0.5) (77 + 54) (0.47)2 = (0.5) (77 + 54) v'2

v' = 0.93 m/s