Question

A horizontal net force F is exerted on anobject at rest. The object starts at x = 0 m andhas a speed of 8.0 after moving 6.0 mmSalong a horizontal frictionless surface. The netforce F as a function of the object's position Xis shown below.Force (N)40302010+k++3Position (m)→6 71245What is the mass of the object?Round answer to 2 significant digits.kg

Answer 1

The force on the object is horizontal.

from graph we can see that the force on the particle after moving 6.0m is,

$F=40\text{N}$

the work done by the force is,

$W=F\times S$

substituting we get,

$\begin{gathered} W=40N\times(6.0-0)m \\ =240\text{J} \end{gathered}$

change in kinetic energy is,

$\frac{1}{2}mv^2-0=240J$

we get this from work energy theorem.

Here v=8.0m/s

so by substituting we get,

$\begin{gathered} \frac{1}{2}m\times8^2=240 \\ m=\frac{240\times2}{64} \\ m=7.5\operatorname{kg} \end{gathered}$

Hence the mass of the object is 7.5kg.