Question

A particle moving along the y-axis has the potential energy u =4y3j, where y is in m. what is the y-component of the force on the particle at y=0 m, 1 m, and 2 m?

Answer 1

Potential energy is given as

$U = 4y^3$

now as we know that force is related by potential energy by the formula

$F = - \frac{dU}{dy}$

So it is gradient of energy with position in Y

$F = - \fracd(4y^3}{dy}$

$F = -12y^2 \hat j$

Now at y = 0

$F = 0 N$

at y = 1

$F = - 12*1^2$

$F = - 12 N$

at y = 2

$F = - 12*2^2$

$F = - 48 N$

so above is the forces at given positions

Answer 2

The force on the particle at

$y=0\;\text{is} \;0\;\text{N}\\\\y=1\;\text{is} \;12\;\text{N}\\\\y=2\;\text{is} \;48\;\text{N}$

Explanation:

Given information:

The potential energy: $U=4y^3$

Now, as we know that force:

$F=-\frac{dU}{dy}$

So, the gradient of energy with position y

$F=-4y^3dy\\F=-12y^2j$

At, y=0

$F=0\;N$.

At, y=1

$F=-12+1\\F=-12 \;N$

At, y=2

$F=-12\times 4\\F=-48\;\text{N}$

Hence , The force on the particle at

$y=0\;\text{is} \;0\;\text{N}\\\\y=1\;\text{is} \;12\;\text{N}\\\\y=2\;\text{is} \;48\;\text{N}$

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