Question

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?

Answer 1

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,

`W=mg\text{ }`

Then, the mass of the car will be

`\begin{gathered} m=(W)/(g) \\ =(8707)/(9.8) \\ =888.46\text{ kg} \end{gathered}`

The acceleration, a can be calculated by the formula

`\begin{gathered} v^2-u^2=2as \\ a=(v^2-u^2)/(2s) \end{gathered}`

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

`\begin{gathered} a=((5.84)^2)/(2*16.73) \\ =(34.10)/(33.46) \\ =1.019m/s^2 \end{gathered}`

The force will be

`\begin{gathered} F=\text{ ma} \\ =888.46*1.019 \\ =\text{ 905.34 N} \end{gathered}`

Thus, the force is 905.34 N