(a) The block is pulled to a position x i = 6.20 cm from equilibrium and release

Question

(a) The block is pulled to a position xi = 6.20 cm from equilibrium and released. Find the potential energy stored in the spring when the block is 6.20 cm from equilibrium.
J

(b) Find the speed of the block as it passes through the equilibrium position.
m/s

(c) What is the speed of the block when it is at a position xi/2 = 3.10 cm?
m/s

Explanation / Answer

(a) potential energy of the spring is E=kx^2/2=890*0.062^2/2=1.71 J (b) when block passes thorough the equilibrium, spring has zero potential energy, which was transferred to the kinetic energy of the block E=mv^2/2=1.71 J --> v^2=2*1.71/1.7=2 --> v=sqrt(2)=1.41 m/s (c) Total energy = potential + kinetic=1.71 --> kx^2/2+mv^2/2=1.71, when x=0.031m --> mv^2=2*1.71-890*0.031^2=2.56 --> v^2=2.56/1.7=1.5 --> v=sqrt(1.5)=1.23 m/s

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