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