A person and a sled of combined 150 kg ride down a frictionless sled track start

Question

A person and a sled of combined 150 kg ride down a frictionless sled track starting 30 m above the ground. a) At the bottom of the track, what is the speed of the person? The sled hits a long patch of dirt with coefficient of kinetic fiction mu k = 0.4. b) How far does the sled progress into the dirt before stopping? c) How much work is done by friction as the sled comes to a stop? d) The sled and rider lost kinetic energy as they ground to a stop, where did that energy go? The sled track contain a circular loop. What is the maximum radius the loop can have such that the sled and person do not fall off during the ride?

Explanation / Answer

a) Apply energy conservation

final kinetic energy = initial potential energy

(1/2)*m*v^2 = m*g*h

v = sqrt(2*g*h)

= sqrt(2*9.8*30)

= 24.25 m/s

b) acceleration of sled, a = -g*mue_k

= -9.8*0.4

= -3.92 m/s^2

distance travelled before stop, d = (vf^2 - vi^2)/(2*a)

= (0^2 - 24.25^2)/(2*(-3.92))

= 75 m

c) Workdone by friction = change in kinetic energy of sled

= 0.5*m*(vf^2 - vi^2)

= 0.5*150*(0^2 - 24.25^2)

= -44105 J

d) the loss of mechaical energy converted into thermal energy.

extra :

to complete a circle,

v_bottom = sqrt(5*g*R)

so, R = V_bottom^2/(5*g)

= 24.25^2/(5*9.8)

= 12 m

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