Question

A toy car, with a mass of 4kg, is pushed with a force of 15N. If the toy car is in the grass with a coefficient of friction of 0.1 then what is the acceleration? How far does it get pushed in 9s?

Answer 1

Answer:

The acceleration = 2.77 m/s²

The distance = 112.185 m

Step-by-step explanation:

The mass of the toy car, m = 4 kg

The force, F = 15 N

Coefficient of Friction, µ = 0.1

The acceleration, a = ?

The frictional force is calculated as:

`\begin{gathered} F_r=\mu mg \\ F_r=0.1(4)(9.8) \\ F_r=3.92N \end{gathered}`

We can calculate the acceleration using the equivalent force formula:

`\begin{gathered} \sum F=F-F_r \\ ma=15-3.92 \\ 4a=11.08 \\ a=(11.08)/(4) \\ a=2.77m/s^2 \end{gathered}`

The acceleration = 2.77 m/s²

To get the distance covered, use the formula below:

`\begin{gathered} S=ut+(1)/(2)at^2 \\ S=0(9)+0.5(2.77)(9^2) \\ S=112.185m \end{gathered}`

The distance = 112.185 m