Chapter 6, Problem 67A

Skelly the skater is propelled by rocket power. Skelly and the rocket together have a mass of 25 kg. The thrusting force is 100 N and friction is 20 N.

a. What is Skelly’s acceleration?

b. How far does he go in 5 s if he starts from rest?

To determine

Acceleration of person S.

Answer to Problem 67A

Acceleration is, $a=3.2{\text{ m/s}}^2$

Explanation of Solution

Given:

Combined mass of person S and rocket is, $m=25\text{ kg}$

Thrust force, $F=100\text{ N}$

Frictional force, $f=20\text{ N}$

Formula used:

According to Newton’s law.

  $F=ma$

Where, m is mass and a is acceleration.

Calculation:

Net force acting on person S is,

  $\begin{array}{l}{F}_{net}=F-f\\ =100-20\\ =80\text{ N}\end{array}$

Now, substituting the value of F and m and solve.

  $\begin{array}{l}80=25\times a\\ a=\frac{80}{25}\\ =3.2{\text{ m/s}}^2\end{array}$

Conclusion:

Therefore, acceleration of person S is $3.2{\text{ m/s}}^2$ .

To determine

Distance travelled by person S.

Answer to Problem 67A

Distance travelled is, $s=40\text{ m}$

Explanation of Solution

Given :

Time taken $t=5\text{ s}$

Formula used:

According to second equation of motion.

  $s=ut+\frac{1}{2}a{t}^2$

Where, s is distance traveled, u is initial speed, a is acceleration and t is time.

Calculation:

From part (a) $a=3.2{\text{ m/s}}^2$

As he starts from rest therefore initial velocity, $u=0{\text{ m/s}}^2$

Now, substituting the value of u , a and t and solve.

  $\begin{array}{l}s=0\times 2+\frac{1}{2}\times 3.2\times 5^2\\ =40\text{ m}\end{array}$

Conclusion:

Therefore, distance travelled by person S is 40 m.