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An object of mass 2 kg attached to a spring is moved to a distance x = 10 m from its equilibrium position. The spring constant k= 1Nm^(-1) and assume that the surface is frictionless (a) When the mass crosses the equilibrium position, what is the speed of the mass? (b) What is the force that acts on the object when the mass crosses the equilibrium position |
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Answer» Solution :(a) Since the spring force is a conservative force, the total energy is constant. At `x=10m`, the total energy is purely potential `E=U=1/2KX^(2)=1/2xx(1)xx(10)^(2)=50J` When the mass crosses the equilibrium position (x = 0), the potential energy `U=1/2xx1xx(0)=0J` The entire energy is purely kinetic energy at this position `E=KE=1/2mv^(2)=50J` The SPEED `v=sqrt((2KE)/m=sqrt((2xx50)/2)=sqrt(50)ms^(-1)=7.07ms^(-1)` (b) Since the restoring spring force is F LOR, when the object crosses the equilibrium position, it experiences no force. Note that at equilibrium position, the object moves very fast. When object is at `x=+10m` (elongation), the force `F=-kx` `F=-(1)(10)=-10N`. Here the negative sign implies that the force is towards equilibrium ie, towards negative x-axis and when the object is at `x=-10` (compression), it experiences a forces `F=-(1) (-10)=+10N`. Here the positive sign implies that the force points towards positive x-axis. The object comes to momentary rest at `x=10m` EVEN THOUGH it experiences a maximum force at both these points |
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