Chapter 5, Problem 103A

a.

To determine

Maximum possible acceleration that a truck can experience.

Answer to Problem 103A

Maximum acceleration of truck without sliding is $4.9{\text{ m/sec}}^2$ .

Explanation of Solution

Given :

Coefficient of static friction between tree and trailor is $\mu =0.50$

Initial velocity of truck $u=55\text{ km/hr}$

Formula used:

Frictional force is given as,

  $F_f=\mu mg$

Where,

  $\mu$ is the coefficient of static friction.

m is mass of body.

g is acceleration due to gravity.

Normal force is given as,

  $F=ma$

Calculation:

For the truck to be stopped without sliding normal force must be equal to frictional force.

Therefore,

  $\begin{array}{c}F=F_f\\ ma=\mu mg\\ a=\mu g\\ a=0.5\times 9.8\\ =4.9{\text{ m/sec}}^2\end{array}$

Conclusion:

Therefore, maximum acceleration of truck without sliding is $4.9{\text{ m/sec}}^2$

b.

To determine

Minimum stopping distance of truck.

Answer to Problem 103A

Truck cover a minimum distance of 308.67 km before coming to rest.

Explanation of Solution

Given :

Coefficient of static friction between tree and trailor is $\mu =0.50$

Initial velocity of truck $u=55\text{ km/hr}$

Formula used:

According to third equation of motion is,

  $v^2=u^2+2as$

Where,

a is the acceleration.

s is the distance travelled.

u is the initial velocity.

v is the final velocity.

Calculation:

As the truck finally stops therefore final velocity of truck $v=0\text{ m/sec}$

From third equation of motion.

  $\begin{array}{c}{0}^2={55}^2+2\times \left(-4.9\right)s\\ 9.8s={55}^2\\ s=\frac{{55}^2}{9.8}\\ =308.67\text{ km}\end{array}$

Conclusion:

Therefore, truck cover a minimum distance of 308.67 km before coming to rest.