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A 8,707 newton car is initiallyat rest. How much force (inNewton) is required to movethe car by 16.73 meters, with afinal velocity of 5.84 m/s?

Sagot :

Given the weight of the car, W = 8707 N, and the car moves a distance, s= 16.73 m, and final velocity, v = 5.84 m/s

Let the mass of the car be m and acceleration due to gravity g = 9.8 m/s^2

Also, weight is given by the formula,

[tex]W=mg\text{ }[/tex]

Then, the mass of the car will be

[tex]\begin{gathered} m=\frac{W}{g} \\ =\frac{8707}{9.8} \\ =888.46\text{ kg} \end{gathered}[/tex]

The acceleration, a can be calculated by the formula

[tex]\begin{gathered} v^2-u^2=2as \\ a=\frac{v^2-u^2}{2s} \end{gathered}[/tex]

Here, u is the initial velocity, u=0.

[tex]\begin{gathered} a=\frac{(5.84)^2}{2\times16.73} \\ =\frac{34.10}{33.46} \\ =1.019m/s^2 \end{gathered}[/tex]

The force will be

[tex]\begin{gathered} F=\text{ ma} \\ =888.46\times1.019 \\ =\text{ 905.34 N} \end{gathered}[/tex]

Thus, the force is 905.34 N