91Ó°ÊÓ

1. $F_1=3260\mathrm{N}$,
2. $F_2=2200\mathrm{N}$and
3. $a=0.39\mathrm{m}/{\mathrm{s}}^2$
   

The weight of the object is equal to the product of mass and gravitational acceleration. The weight is form of force. Newton’s law states that the force acting on the object is equal to mass times acceleration of the object.

Find weight of the landing craft by using weight equation. We use Newton’s second law to find acceleration.

Formulae:

1. W=mg
2. F =ma

Here,W is weight of the object,m is mass of the object,g is gravitational acceleration,F is force anda is acceleration of the object.

To descend at constant speed the force by thrust will be equal to weight of the craft.

We can take downward as positive direction for simplicity; we apply Newton’s second law,

$$\begin{gathered} ma=mg-F_1 \\ m\left(0\right)=mg-3260 \\ mg=3260\mathrm{N} \\ W=3260\mathrm{N} \end{gathered}$$

Weight of the craft is 3260 N.

For second force there is acceleration $a_1=0.39\mathrm{m}/{\mathrm{s}}^2$.

We can apply Newton’s second law,

$$\begin{gathered} m{a}_1=mg-F_2 \\ m=\frac{mg-F_2}{a_1} \\ =\frac{3260\mathrm{N}-2200\mathrm{N}}{0.39\mathrm{m}/{\mathrm{s}}^2} \\ =2.7×{10}^3\mathrm{kg} \end{gathered}$$

The mass of the craft is $2.7×{10}^3\mathrm{kg}$.

Use the weight of the craft to find the free fall acceleration.

$$\begin{gathered} W=mg \\ g=\frac{W}{m} \\ g=\frac{3260\mathrm{N}}{2.7×{10}^3\mathrm{kg}} \\ =1.2\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Therefore, the free all acceleration is $1.2\mathrm{m}/{\mathrm{s}}^2$.