College Physics by Openstax Chapter 4 Problem 7


(a) If the rocket sled shown in Figure 4.31 starts with only one rocket burning, what is the magnitude of its acceleration? Assume that the mass of the system is 2100 kg, the thrust TT is 2.4×1042.4 \times 10^{4} N, and the force of friction opposing the motion is known to be 650 N. (b) Why is the acceleration not one-fourth of what it is with all rockets burning?


Solution:

Considering the direction of motion as the positive direction, we are given the following: T=2.4×104 NT=2.4 \times 10^4 \ \text{N}, f=650 Nf=-650 \ \text{N}, and mass, m=2100 kgm=2100 \ \text{kg}.

Part A. The magnitude of the acceleration can be computed using Newton’s Second Law of Motion.

ΣF=ma2.4×104 N650 N=2100 kg×a23350=2100a233502100=2100a2100a=233502100a=11 m/s2  (Answer)\begin{align*} \Sigma F & =ma \\ 2.4\times 10^4 \ \text{N}-650 \ \text{N} & = 2100 \ \text{kg}\times a \\ 23350 & = 2100 a \\ \frac{23350}{2100} & = \frac{\cancel{2100} a}{\cancel{2100}} \\ a & = \frac{23350}{2100} \\ a & = 11 \ \text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B. The acceleration is not one-fourth of what it was with all rockets burning because the frictional force is still as large as it was with all rockets burning.   (Answer)\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)


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