Need to know how to do these questions: Any physics gurus please helps:

haha yo question exam ma aaunewala cha hoo babu???...lol..

mass*acceleration=mass*gsin(45)-coeff*mass*gcos(45)
cancel mass and get your answer with proper substitution.

यो त सार्है सजिलो भएन र गुरुकै किन खोजी गर्नुपर्यो चेलाले पनि जान्लान नि। हुन त धेरै भो यस्ता हिसाब नदेखेको , त्यही पनि ट्राइ मारुम है त?
१) F= W*Sin40
W*Sin40-f=Ma
W*Sin40-uN=Ma
W*Sin40-uWcos40=Ma
W = weight,u=coeff frction, a=acceleration, M=mass
answer should be around 0.76 आंफै गरु है।

हो कि हो कि जस्तो लाग्छ यो पनि।
Cforce=W-N
mv^2/r=W-N
N=0 भए गाडी खस्छ
mv^2/r=mg-0
m=mass v=velocity,r=radius,g=earth grav
answer should be around 7.081 आंफै गरु है।

nepalean chai guru nai ho rahecha...... Thanks dude.......

For the Second part, I think answer should be as follows:
f = (V^2/ 15R) - 0.01 e
where,
f = Side friction, which depends on road type, condition and speed of vehicle
V = Speed in mph, R = Radius in ft
e = Road super elevation
Since f depends on speed as well, the solution should be in quadratic form.
Suppose, e =0 (no superelevation)

Speed (mph)             f-value
10                             0.21
20                             0.21
30                             0.18
40                             0.16
50                             0.14
60                             0.13
70 & more                 take 0.10

So, pick a f value, calculate speed and check if it matches for your speed taken for that f-value.
(I forgot equation for f -value)

For Example:
e =0, f=0.17, R = 558 ft (170m)
Hence, V= 37 mph, and for this speed f=0.17
I guess, your answer should be 37 mph.

My common sense says

a)  Coeff.  * g cos 40 = g sin 40 - 0.55

b) V_max ^2 / R = g   (they're talking about the top point)

Newton's First Law: Qualitative definition of Force. If no NET EXTERNAL FORCE acts on an object, the object will neither change direction nor its speed, nor both.

Start: What is the net external force in your problem?

What are the forces acting on the object? Gravity and Friction.

Check: Are they acting on the same direction? Yes or No.

In this case: No.

If No, then are they acting perpendicular to each other? No.

If No, then the component of the Gravity and Friction in the direction of motion (in this case sliding). Component of gravity parallel to direction of motion = Weight * sin_theta (where theta is the  angle the motion surface makes with horizontal).

Friction is acting parallel to the surface of the motion, but always acts against (opposite) the motion. So, it is mu_k(coef. of kinetic friction) * component of force normal to the surface of motion. 

The component of force normal to the surface of motion = Weight * cos_theta (where theta is the  angle the motion surface makes with horizontal).

Net external force (=net driving force) = difference between aforemention two forces = Gravity component in the direction of motion - friction force on the surface of motion(note it always acts against motion) = W * sin_theta - mu_k* W*cos_theta

Now, what is second law of motion (Newton's second law). Force = mass * acceleration.  Similarly, apply newton's third law of motion: action (= net driving force)  = reaction (i.e. mass * acceleration)

So, mass* acceleration = driving force =  W * sin_theta - mu_k* W*cos_theta  (replace W by mass*g )

So, mass* acceleration = mass*g * sin_theta - mu_k* mass*g *cos_theta

cancel mass from the equation because it is common.

Solve this equation for mu_k (coefficient of friction) = (g * sin_theta - acceleration )/ (g *cos_theta)

You are done. Note, if you get mu_k negative, then what is your interpretation? As I said above, friction always acts against the motion. You must think it and make you interpretation correctly. Friction is a non-conservative force. If you get negative coefficient of friction, the net driving force is not enough to counter the friction, and therefore, there is acceleration, but decceleration. If the object is not in motion as an initial condition, then you will know know kinetic friction. Your answer is "not enough information". Period.