2. 2/4 points| Previous Answers My Notes The diagram below is a top-down view of

Question

2. 2/4 points| Previous Answers My Notes The diagram below is a top-down view of two children pulling a 10.0-kg sled along the snow. The first child exerts a force of Fi 10 N at an angle 01-45 counterclockwise from the positive x direction. The second child exerts a force of F2-8 N at an angle 2 = 30° clockwise from the positive x direction. Fi (a) Find the magnitude (in N) and direction of the friction force acting on the sled i it moves with constant velocity. magnitude 14.33 direction (counterclockwise from the +x-axis) 192.37 What is the coefficient of kinetic friction between the sled and the ground? 1.237 What is the relationship between friction force and normal force? (b) (c) What is the magnitude of the acceleration (in m/s2) of the sled if Fi is doubled and F2 is halved in magnitude? 1.85 Doubling or halving a force will not necessarily have the same effect on the resultant. m/s

Explanation / Answer

b) we know, fk = mue_k*N

fk = mue_k*m*g

==> mue_k = fk/(m*g)

= 14.33/(10*9.8)

= 0.146

b) F1' = 2*F1 = 2*10 = 20 N
F2' = F2/2 = 8/2 = 4 N

Fnet = sqrt(F1'^2 + F2'^2 + 2*F1'*F2'*cos(theta1+theta2))

= sqrt(20^2 + 4^2 + 2*20*4*cos(45 + 30))

= 21.4 N

a = Fnet/m

= 21.4/10

= 2.14 m/s^2

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