When the 150-lb skier is at point A he has a speed of 5 ft/s. When the skier rea

Question

When the 150-lb skier is at point A he has a speed of 5 ft/s. When the skier reaches point B on the smooth slope, the slope follows the cosine curve shown. Suppose that y(x)=62cos(/100 * x)ft, where x is in ft. Neglect friction and air resistance. (Figure 1).

a) Determine the skier's speed when he reaches point B on the smooth slope.

b) What is the normal force on the skier's skis at B?

c) What is the skier's rate of increase in speed at B?

When the 150-lb skier is at point A he has a speed of 5 ft/s. When the skier reaches point B on the smooth slope, the slope follows the cosine curve shown. Suppose that 62cosf Figure 1) r ft, where z is in ft. Neglect friction and air resistance

Explanation / Answer

a)

at x = 0; y = 62ft

at x = 35 ; y = 62cos(180 x 35/100) = 28.15 ft

u = 5 ft/s ; v = ? ; a = g = 32 ft/s^2

using 3rd kinematical eqn

v^2 = u^2 + 2as

= 5^2 + 2 x 32 x (62-28.15)

v = 46.81 ft/s

b)

dy/dx = -62 x 3.14/100 x sin(180X/100)

= -1.735

but dy/dx = tan@

@ = tan-1(-1.735) = -60 deg

Normal reaction = Wcos@ = 150cos60 = 75 lbs

c)

a = gsin@ = 32 sin60 = 27.7 ft/s^2

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