A particle moving along the y-axis is in a system with potential energy $U=4y^{3}J$, where y is in m. What is the y-component of the force on the particle at $y=0m,1m,and2m$?

The potential energy equation is given as

$U=4y^{3}J$where y is in 'm'

To find: y-component of the force at$y=0m,1m$and$2m$.

The force exerted on a mass item that causes it to alter velocity.

Potential energy of a motion equation along y-axis,

$U=4y^{3}J$

On the potential energy diagram, the equation gives the y-component of force,

localid="1649680949546" $F_y=-\frac{dU}{dy}J/m$

The slope of the potential energy diagram is the y-component of force.

Particle slope, on substituting the potential energy equation, we get

localid="1649680968976" $$\begin{gathered} F_y=-\frac{d}{dy}\left(4y^3\right).....\left(1\right) \\ =-12y^2 \end{gathered}$$

At localid="1649680998077" $y=0m$, the y-component of force is,

localid="1649681022888" $$\begin{gathered} F_{y=0m}=-12{\left(0\right)}^2 \\ =0N \end{gathered}$$

At localid="1649681008000" $y=1m$, the y-component of force is,

localid="1649681029255" $$\begin{gathered} F_{y=1m}=-12{\left(1\right)}^2 \\ =-12N \end{gathered}$$

At localid="1649681014999" $y=2m$, the y-component of force is,

localid="1649681036714" $$\begin{gathered} F_{y=2m}=-12{\left(2\right)}^2 \\ =-48N \end{gathered}$$

Hence,

y-component of force at localid="1647868600107" $y=0m$ localid="1649681046695" $=0N$.

y-component of force at localid="1647868612840" $y=1m$ localid="1649681054130" $=-12N$.

y-component of force at $y=2m$localid="1649681061631" $=-48N$.