PHYSICS COLLEGE

Before 1960, people believed that the maximum attainable coefficient of static friction for an automobile tire on a roadway was ?s = 1. Around 1962, three companies independently developed racing tires with coefficients of 1.6. This problem shows that tires have improved further since then. The shortest time interval in which a piston-engine car initially at rest has covered a distance of one-quarter mile is about 4.43 s. (A) Assume the car's rear wheels lift the front wheels off the pavement as shown in the figure above. What minimum value of ?s is necessary to achieve the record time?

Answers

Answer 1

Answer:

Answer:

4.18

Explanation:

Givens

The car's initial velocity $v_(i)$ = 0 and covering a distance Δx = 1/4 mi = 402.336 m in a time interval t = 4.43 s.

Knowns

We know that the maximum static friction force is given by:

$f_(s_max) =$ μ_s*n (1)

Where μ_s is the coefficient of static friction and n is the normal force.

Calculations

(a) First, we calculate the acceleration needed to achieve this goal by substituting the given values into a proper kinematic equation as follows:

Δx= $v_(i) +(1)/(2) at^2$

a=41 m/s

This is the acceleration provided by the engine. Applying Newton's second law on the car, so in equilibrium, when the car is about to move, we find that:

$f_(y)=n-mg=0\n n=mg\nf_(x)=F-f_(s,max) =0\n f_(s,max)=F=ma\n$

Substituting (3) into (1), we get:

$f_(s,max)=$ μ_s*m*g

Equating this equation with (4), we get:

ma= μ_s*m*g

μ_s=a/g

=4.18

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What is the force (in Newtons, 1 Newton = 1Kgm/s2) required to accelerate a 1500 Kg car to 3 m/s2?

A rocket sled accelerates at a rate of 49.0 m/s2 . Its passenger has a mass of 75.0 kg. (a) Calculate the horizontal component of the force the seat exerts against his body. Compare this with his weight using a ratio. (b) Calculate the direction and magnitude of the total force the seat exerts against his body.

Answers

Explanation:

It is given that,

Mass of the passenger, m = 75 kg

Acceleration of the rocket, $a=49\ m/s^2$

(a) The horizontal component of the force the seat exerts against his body is given by using Newton's second law of motion as :

F = m a

$F=75\ kg* 49\ m/s^2$

F = 3675 N

Ratio, $R=(F)/(W)$

$R=(3675)/(75* 9.8)=5$

So, the ratio between the horizontal force and the weight is 5 : 1.

(b) The magnitude of total force the seat exerts against his body is F' i.e.

$F'=√(F^2+W^2)$

$F'=√((3675)^2+(75* 9.8)^2)$

F' = 3747.7 N

The direction of force is calculated as :

$\theta=tan^(-1)((W)/(F))$

$\theta=tan^(-1)((1)/(5))$

$\theta=11.3^(\circ)$

Hence, this is the required solution.

Final answer:

The horizontal component of the force the seat exerts against the passenger's body is 3675 N. The ratio of this force to the passenger's weight is 5. The total force the seat exerts has a magnitude of 3793 N.

Explanation:

(a) To calculate the horizontal component of the force the seat exerts against the passenger's body, we can use Newton's second law, which states that force is equal to mass times acceleration. In this case, the mass of the passenger is 75.0 kg and the acceleration of the rocket sled is 49.0 m/s2. So the force exerted by the seat is:

Force = mass * acceleration

Force = 75.0 kg * 49.0 m/s2

Force = 3675 N

Now let's compare this force to the passenger's weight. The weight of an object is given by the formula:

Weight = mass * gravitational acceleration

Weight = 75.0 kg * 9.8 m/s2

Weight = 735 N

To find the ratio, we divide the force exerted by the seat by the weight of the passenger:

Ratio = Force / Weight

Ratio = 3675 N / 735 N

Ratio = 5

(b) The total force the seat exerts against the passenger's body has both a horizontal and vertical component. The direction of the total force is the same as the direction of the acceleration of the rocket sled. The magnitude of the total force can be found using the Pythagorean theorem:

Total Force = √(horizontal component2 + vertical component2)

Total Force = √(36752 + 7352)

Total Force = 3793 N