Question

A luge and its rider, with a total mass of 99 kg, emerge from a downhill track onto a horizontal straight track with an initial speed of 34 m/s. If a force slows them to a stop to a stop at a constant rate of 1.8 m/s^2, (a)What magnitude F is required for the force, (b)What distance d do they travel while slowing, and(c) What work W is done on them by the force? What are (d) F, (e) d and (f) W if they, instead slow at 3.6 m/s^2

Answer 1

Since, Force required to stop is given by,

F = m*a

here, m = mass of rider = 99 kg

a = de-acceleration = 1.8 m/sec^2

So, F = 99*1.8

F = 178.2 N

b.)

by kinematics law,

v^2 = u^2 + 2*a*S

here, v = final velocity = 0 m/sec.

u = initial velocity = 34 m/sec.

a = -1.8 m/sec.^2

S = (v^2 - u^2)/(2*a) = (0^2 - 34^2)/(2*(-1.8))

S = 321.11 m

c.)

Work done by force = W = F*S*cos(x)

here, x = angle between force and distance vector = 180 deg

So, W = 178.2*321.11*cos(180 deg)

W = -5.72*10^4 J

d.)

Now, a = 3.6 m/sec^2

F = 99*3.6

F = 356.4 N

e.) Since, distance travel is given by ,

S = (v^2 - u^2)/(2*a) = (0^2 - 34^2)/(2*(-3.6))

S = 160.56 m

f.)

also, W = 356.4*160.56*cos(180 deg)

W = -5.9*10^4 J

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