91影视

The acceleration of the object at constant force F is given by$a=8m/s^2$.

Since force is directly proportional to the acceleration of the object by Newton's Second law, it can be written as

$$\begin{gathered} F=ma \\ F鈭�a \end{gathered}$$

Consider the initial force to be F and initial acceleration be a.

If the force F' is doubled, then the acceleration a' is also doubled.

Therefore, the new acceleration is calculated as :

$$\begin{gathered} \frac{F}{F^{\text{'}}}=\frac{a}{a^{\text{'}}} \\ \frac{F}{2F}=\frac{8}{a^{\text{'}}} \\ {a}^{\text{'}}=16m/s^2 \end{gathered}$$

The formula to calculate the force acting on the object is given by

$F=ma$

In terms of relation of acceleration and mass of the object, it can be written as

$$\begin{gathered} a=\frac{F}{m} \\ a鈭�\frac{1}{m} \end{gathered}$$

Consider the original acceleration be $a=8m/s^2$and mass be m.

When the mass is doubled, let the final mass be m' and the acceleration be a' respectively.

The new acceleration of the object when the mass is doubled is given by

$$\begin{gathered} \frac{a^{\text{'}}}{a}=\frac{m}{m^{\text{'}}} \\ {a}^{\text{'}}=\frac{m}{2m}脳8m/s^2 \\ =4m/s^2 \end{gathered}$$

Consider the original force, F, mass be m and acceleration of the object be a=8m/s².

The new force be F' =2F and mass be m'=2m respectively.

The new acceleration when the force and the object鈥檚 mass are both doubled can be calculated as

$$\begin{gathered} F=ma \\ a=\frac{F}{m} \\ \frac{a^{\text{'}}}{a}=\frac{{\displaystyle \raisebox{1ex}{$F^{\text{'}}$}\!\left/ \!\raisebox{-1ex}{$m\text{'}$}\right.}}{{\displaystyle \raisebox{1ex}{$F$}\!\left/ \!\raisebox{-1ex}{$m$}\right.}} \\ =\frac{{\displaystyle \raisebox{1ex}{$2F$}\!\left/ \!\raisebox{-1ex}{$2m$}\right.}}{{\displaystyle \raisebox{1ex}{$F$}\!\left/ \!\raisebox{-1ex}{$m$}\right.}} \\ {a}^{\text{'}}=a \\ =8m/s^2 \end{gathered}$$

Consider the original force, F, mass be m and acceleration of the object be a=8m/s².

The new force be F' =2F and mass be m'=m/2 respectively.

The new acceleration when the force is doubled and the object鈥檚 mass is halved can be calculated as:

$$\begin{gathered} F=ma \\ a=\frac{F}{m} \\ \frac{a^{\text{'}}}{a}=\frac{{\displaystyle \raisebox{1ex}{$F^{\text{'}}$}\!\left/ \!\raisebox{-1ex}{$m\text{'}$}\right.}}{{\displaystyle \raisebox{1ex}{$F$}\!\left/ \!\raisebox{-1ex}{$m$}\right.}} \\ =\frac{{\displaystyle \raisebox{1ex}{$2F$}\!\left/ \!\raisebox{-1ex}{$\left({\displaystyle \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)$}\right.}}{{\displaystyle \raisebox{1ex}{$F$}\!\left/ \!\raisebox{-1ex}{$m$}\right.}} \\ {a}^{\text{'}}=4a \\ =32m/s^2 \end{gathered}$$