# Problem: The position of a 2.2-kg mass is given by x(t) = (2t 3 - 4t 2) m, where t is in seconds.(a) What is the net horizontal force on the mass at t = 0  s?(b) What is the net horizontal force on the mass at t = 1  s?

###### FREE Expert Solution

In this problem, we are required to determine the magnitude of the force on an object given its position as a time-varying function

Since we have the position as a function of time, we know this is a calculus problem.

A diagram like this one can help you remember the relationships between the variables:

$\mathbit{P}\begin{array}{c}{\mathbf{←}}\\ {\mathbf{\to }}\end{array}\underset{\frac{\mathbit{d}}{\mathbit{d}\mathbit{t}}}{\overset{{\mathbf{\int }}{\mathbit{d}}{\mathbit{t}}}{\mathbit{V}}}\begin{array}{c}{\mathbf{←}}\\ {\mathbf{\to }}\end{array}\underset{\mathbit{F}\mathbf{=}\mathbit{m}\mathbit{a}}{\mathbit{A}\mathbf{,}\mathbit{F}}$

Since we're looking for the magnitude of the force exerted on the object, we'll have to find an expression for acceleration, a(t). The a(t) expression is obtained by differentiating the position function, x(t), twice.  Then, we'll multiply the a(t) function by the mass, m, to get the force expression.

The steps needed to solve this problem straightforward:

1. Differentiate the position function, x(t), twice to get the acceleration function a(t). Remember, differentiating the position function, x(t), once gives the velocity function, v(t).
2. Use the equation ma to find a function for the force
3. Calculate the force at a specific time of interest (if required).

In Step 1, we'll need to remember the power rule of differentiation:

$\overline{)\frac{\mathbf{d}}{\mathbf{dt}}\mathbf{\left(}{\mathbf{t}}^{\mathbf{n}}\mathbf{\right)}{\mathbf{=}}{{\mathbf{nt}}}^{\mathbf{n}\mathbf{-}\mathbf{1}}}$

###### Problem Details

The position of a 2.2-kg mass is given by x(t) = (2t 3 - 4t 2) m, where t is in seconds.
(a) What is the net horizontal force on the mass at t = 0  s?
(b) What is the net horizontal force on the mass at t = 1  s?