A block having mass m and charge +Q is connected to aspring having constant k .

Question

A block having mass m and charge +Q is connected to aspring having constant k. The block lies on a frictionless horizontal track, and the system isimmersed in a uniform electric field of magnitude E, directed as shown inthe figure. If the block is released from rest when thespring is unstretched (at x = 0), (a) by what maximum amount does thespring expand? (b) What is the equilibrium position of the block? (c) Show that theblock’s motion is simple harmonic, and determine its period. (d) What If? Repeatpart (a) if the coefficient of kinetic friction between block and surface is k. A block having mass m and charge +Q is connected to aspring having constant k. The block lies on a frictionless horizontal track, and the system isimmersed in a uniform electric field of magnitude E, directed as shown inthe figure. If the block is released from rest when thespring is unstretched (at x = 0), (a) by what maximum amount does thespring expand? (b) What is the equilibrium position of the block? (c) Show that theblock?s motion is simple harmonic, and determine its period. (d) What If? Repeatpart (a) if the coefficient of kinetic friction between block and surface is Muk.

Explanation / Answer

a) force on charge due to field is QE, this force will changeinto restoring force in spring when the spring will stop aftercovering(stretching) the distance "x"(say) Now restoring force is -kx So -kx=QE x=QE/k (Don't consider -sign as it just shows that restoringforce is opposite to QE) b) The equlibrium position is x/2 which will be QE/2k c) From a) QE=-kx dividing by m to get acceleration, So acceleration A=F/m=QE/m=-kx/m A=-(k/m)x Now k/m is constant, so acc is proportional to distance frommean position and is directed towards mean position(due to - sign)so motion is SHM(simple harmonic motion) To find time period put 2=k/m =2/T so 42/T2=k/m Solve for T which would be =2m/k d) for part d, restoring force would beQE-kmg Now take distance covered as x' So kx'=QE-kmg Solve for x' by dividing both sides by k could you check the answers please? So kx'=QE-kmg Solve for x' by dividing both sides by k could you check the answers please?

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