Two-dimensional force problems.

1.A 56-kg person on skis starts from rest down a hill sloped at 34° from the horizontal. The coefficient of friction between the skis and the snow is 0.19. After the skier had been moving for 5.0 s, the friction of the snow suddenly increased and made the net force on the skier zero.

(a)What is the new coefficient of friction?

(b)How fast would the skier now be going after skiing for another 5.0 s?

A 60-kg person on skis is going down a hill sloped at 30° from the horizontal. The coefficient of friction between the skis and the snow is 0.20. What would be the magnitude of the acceleration?

Joe wishes to hang a sign weighing 740 N so that cable A, attached to the store, makes a 30.0° angle, as shown below. Cable B is horizontal and attached to an adjoining building. What is the tension in cable B?

Hello

1)
friction force = μmgcos34
gravitational force = mgsin34
net force = mg(sin34 - μcos34)
acceleration = F/m = g(sin34 - μcos34)
v = at = g(sin34 - μcos34)*5 = 19.7 m/s (= velocity at the end of the first 5 seconds)

now friction force = gravit. force:
mgsin34 = μmgcos34
μ = sin34/cos34
μ = 0.674 <-- new coefficient

If the net force is = 0, then the skier keeps the velocity as at the end of the first 5 seconds
= 19.7 m/s
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2)
as above:
Fnet = mgsin30 - μmgcos30
a = Fnet/m = g(sin30 - μcos30)
a = 3.2 m/s
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3)
the 30° angle is with the horizontal?

draw a parallelogram of forces, the upright diagonal is the weight of the sign, the sides are the cables, and parallels to the cables. Find the magnitude of the horizontal sides in force units:
F = 740 N/tan30 = 1281.7 N (= force in cable B)

Regards