A horizontal spring exerts a force F = (− kx , 0, 0) that is proportional to its deformation in the axial or x direction. The work of this spring on a body moving along the space curve s ( t ) = ( x ( t ), y ( t ), z ( t )), is calculated using its velocity, v = ( v x , v y , v z ), to obtain W = ∫ 0 t F ⋅ v d t = − ∫ 0 t k x v x d t = − ∫ 0 t k x d x d t d t = ∫ x ( t 0 ) x ( t ) k x d x = 1 2 k x 2 {\displaystyle W=\int _{0}^{t}\mathbf {F} \cdot \mathbf {v} \mathrm {\,} {d}t=-\int _{0}^{t}kxv_{x}\mathrm {\,} {d}t=-\int _{0}^{t}kx{\frac {\mathrm {d} x}{\mathrm {d} t}}dt=\int _{x(t_{0})}^{x(t)}kx\ \mathrm {d} x={\frac {1}{2}}kx^{2}} For convenience, consider contact with the spring occurs at t = 0, then the integral of the product of the distance x and the x -velocity, xv x , is x 2/2. The function U ( x ) = 1 2 k x 2 , {\d...
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