I need some help with these two questions, the one on the left asks \"Sam, who\'

Question

I need some help with these two questions, the one on the left asks "Sam, who's mass is 64.0 kg, takes off down a 50.0m high, 11 degree slope on his jet powered skis. The skis have a thrust of 220 N. Sams speed at the bottom is 38.0 m/s" what is the coefficient of kinetic friction of his skis on snow?
C & https//session.masteringphyc Phys201-2016-5pring MasteringPhysics Chapters 586-Google Chrome ps//session ys201-2016-Spring Problem 6 28 What in the Espress you Earn t roblem 643 Sam, whose mass is 64 0 kg sakes off don a 50 0 an high. 110 slope on is jetl powered skis The skis have a thst of 220 N Sam's speed a the bothom is 380 m/s What is the coofficient of kinec iction of hin skis on T 1033 Incoerect, Try Again; 5 attempts remaining . 13 Incorrect Try Again; 5 attempts remaining

Explanation / Answer

problem no; 6.43

net force on sam
F(net) = ma (down) = T (thrust, down along) + mg sin 11 (down along) - µk [mg cos 11] (now down)

ma = 220 + mg sin 11 - µk [mg cos 11]
a = [220/64] + 9.8* sin 11 - µk [9.8 cos 11]
a = 3.43 + 1.86 - µk [9.61]
a = acceleration down = 5.29 - µk [9.61] ---------- (1)
----------------------
v (bottom) = 38 m/s
u (top) =0
L = length of incline= 50 / sin 11 = 262.04 m
v^2 = u^2 + 2 a L = 0+ 2 a L
a = 38*38/2*262.04 = 2.75 m/s^2
put in (1)
2.75 = 5.29 - µk [9.61]
µk [9.61] = 2.54
µk = coeff kinetic friction = 0.264

problem no; 6.28

Start by drawing three arrows representing all the forces present.

Draw your beam. Place a dot at the center of your beam.

Now draw your first arrow pointing away from the dot at 20 degrees from Y axis angled to the left.

Then you draw your second arrow pointing away from the dot angled at 20 degrees from Y axis.

Now the third arrow will be point directly downwards from the dot.

All tails of the arrows should be connected to the dot.

This is your static force diagram.

1. Now you realise that the two top cables are working against each other.
2. Now you realise that the two top cables are working together against the bottom arrow.

Once you make those assertions, you can start constructing your static equations.

The bottom arrow is directly on the Y axis so we don’t have to make any angle conversions.

The other two arrows are diagonal, so they have an upwards and sidewards component.

I will help you out by construct the static equation for the vertical situations:

Note that T1 is force in 20 degree cable
Note that T2 is force in 30 degree cable
Note that W is weight of beam

For the vertical forces to be equal:

T1 cos 20 + T2 sin 30 = W
T1 cos 20 + T2 sin 30 = 1100*9.81 (converting to newtons using gravity)

now for the horizontal situation. Remember, the bottom arrow doesn’t have any sidewards force so the only arrows we consider are the top two working against each other.

T1 sin 20 = T2 cos 30

Now you have simultaneous equations which you can solve for:

T1 cos 20 + T2 sin 30 = 1100*9.81
T1 sin 20 = T2 cos 30  

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