Newton's laws and gravitation. help pls.

Question

1. A particular race car can cover a quarter-mile track (402m) in 6.40s starting from a standstill. Assuming the acceleration is constant, how many “g’s” does the driver experience? If the combined mass of the driver and race car is 485kg, what horizontal force must the road exert on the tires?

2. A sports car of mass 950kg (including the driver) crosses the rounded top of a hill (radius = 95m) at 22m/s. Determine

a) the normal force exerted by the road on the car,

b) the normal force exerted by the car on the 72kg driver, and

c) the car speed at which the normal force on the driver equals zero.

3. Two blocks of masses m1 and m2 are connected to each other and to a central post by cords. They rotate about the post at a frequency f (revolutions per second) on a frictionless horizontal surface at distances r1 and r2 from the post. Derive an algebraic expression for the tension in each segment of the cord.

Answer 1

I am giving you the solution with the hope that you will use it to really learn things and not only to get your homework done. If you learn, then I assure you that you will find your course very interesting and you will enjoy solving problems yourself.

1. Let acceleration = a

Displacement s \ 402 m

Time t = 6.4 s

Initial velocity u = 0

s = ut + 1/2 at^2

402 = 0 + 1/2 * a * 6.4^2

Or, 402 = 1/2 * a * 41

Or, a = 402 * 2/41 = 19.6 m/s^2

a/g = 19.6/9.8 = 2

The driver experiences 2 g's.

Mass m = 485 kg

Force = ma = 485 * 19.6 N = 9506 N

2. a. Consider car + driver as system.

Mass m = 950 kg

Radius r = 95 m

Speed v = 22 m/s

Let N = normal force by road on car. This force is vertically upward.

Weight mg acts downward.

Therefore, centripetal force = mg - N

But centripetal force = mv^2/r,

Therefore, mg - N = mv^2/r

Or N = mg - mv^2/r

Now you can calculate N.

b) Solve the same way as in (a) with one difference. For m, use driver's mass = 72 kg

c) N = mg - mv^2/r

If N = 0, then

mg = mv^2/r

Dividing by m,

g = v^2/r

Or, v = sqrt(rg)

Calculate v

3. No. of revoluions in 1 second = f

Distance covered in 1 revolution = 2 * pi * r, where r is radius of circular path

Therefore, distance covered in 1 second = 2 * pi * r * f

Or, speed v = 2*pi*r*f

Tension acts as centripetal force.

Therefore,

T = mv^2/r, where m is the mass of the object.

Therefore, tension in the part connecting to m1

= m1v^2/r1 = m1 * (2*pi*r1*f)^2/r1

= m1 * 4 * pi^2 * r1^2 * f^2/r1

= 4m1 * pi^2 * r1 * f^2

Tension in the part connected to m2 = = 4m2 * pi^2 * r2 * f^2

[e-mail]