Question

The position of a 2.0 kg mass is given by x = 12t 3 - 3t 2 2 m, where t is in seconds. What is the net horizontal force on the mass at

(a) t = 0 s and

(b) t = 1 s?

Short answer

Net horizon force acting on the mass changes from

a. $F=-12N$at $t=0$s

b.$F=12N$at$t=1s$

Step-by-step solution

Content Introduction

The force acting on the mass are$F=-12Natt=0s$and$F=12Natt=1s$

Explanation (Part a)

mass of object $m=2.0kg$

Equation position role="math" localid="1647602349628" $x=2t3-3t2m$

Formula used for velocity

$$\begin{gathered} v=dxdt \\ v=ddt(2t3-3t2) \\ v=2ddt\left(t3\right)-3ddt\left(t2\right) \\ v=2\left(3t2\right)-3\left(2t1\right) \\ v=6t2-6t \end{gathered}$$

For acceleration

$$\begin{gathered} a=dvdt \\ a=ddt(6t2-6(t)-6ddt(t) \\ a=12t-6 \\ Att=0s \\ a=6m/s^2 \end{gathered}$$

For net horizon force F

$$\begin{gathered} F=ma \\ F=2\times (-6) \\ F=-12N \end{gathered}$$

Explanation (Part b)

When for $t=1s$

We are given

mass of object $m=2.0kg$

Equation position $x=2t3-3t2m$

For velocity, formula used is

$$\begin{gathered} v=dxdt \\ v=ddt(2t3-3t2) \\ v=2ddt\left(t3\right)-3ddt\left(2t\right) \\ v=2\left(3t2\right)-3\left(2t1\right) \\ v=6t2-6t \end{gathered}$$

For acceleration

$$\begin{gathered} a=dvdt \\ a=ddt(6t2-6t) \\ a=6ddt\left(t\right) \\ a=12t-6 \\ Att=1s \\ a=6m/s^2 \end{gathered}$$

For net horizon force F

role="math" localid="1647602742699" $$\begin{gathered} F=2\times \left(6\right) \\ F=12N \end{gathered}$$