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A 7kg block on a horizontal frictionless surface is attached to a light spring (force constant= 1.2 kN/m). The block is initially at rest at its equilibrium position, it has a speed of .8m/s. How much work in J is done on the block by the force P as the block moves the 8 cm?

2006-11-11 11:22:18 · 1 answers · asked by Shane W 1 in Science & Mathematics Physics

1 answers

Frictionless flat surface, so no angles to deal with.

But cant you just use the final velocity, and KE= 1/2 * m *v*v ?

KE = 0.5*(m)*v2

KE = (0.5) * (7kg) * (0.8 m/s)2

KE = 2.240 Joules (kg*m*m)/(s*s)

2006-11-11 11:32:29 · answer #1 · answered by Austin Semiconductor 5 · 0 0

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