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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

A horizontal net force F is exerted on anobject at rest. The object starts at x = 0 m-example-1
User Eran Meiri
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1 Answer

5 votes

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* S

substituting we get,


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

change in kinetic energy is,


(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} (1)/(2)m*8^2=240 \\ m=(240*2)/(64) \\ m=7.5\operatorname{kg} \end{gathered}

Hence the mass of the object is 7.5kg.

User Conway
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