A ball of unknown mass m is tossed straight up with initial speed v. At the mome

Question

A ball of unknown mass m is tossed straight up with initial speed v. At the moment it is released, the ball is a height h above a spring-mounted platform. The ball rises, peaks, and falls back toward the platform, ultimately compressing the spring a maximum distance d from its relaxed position. Assume that the spring is perfectly ideal with spring constant k, and that the mass of the spring and platform is negligible. In previous steps of this tutorial, you wrote down expressions A for the total initial energy of the system and the total final T energy of the system. Since the total energy of the system is v constant, the expressions for the initial and final system energy can be set equal to each other. The resulting equation can be solved for the unknown quantity. The original question asked you to find the mass m of the ball. Instead of finding the mass m now, solve the energy " equation for the initial height h of the ball in terms of m, v, d, k, and g.

Explanation / Answer

lets consider the uncompressed spring height (i.e. the surface of the platform when spring is relaxed)

be the referrence point for potential energy.

i.e. at this level, potential energy =0

then total initial energy of the system

=initial potential energy+initial kinetic energy

=m*g*h+0.5*m*v^2

now, in the final position, spring is compressed by d.

and the speed of the object is 0.

then total final energy=

final potential energy+final kinetic energy

=final potential energy of the object+potential energy of the spring+kinetic energy

=-m*g*d+0.5*k*d^2+0

=0.5*k*d^2-m*g*d

using conservation of energy principle and equating initial and final energy:

m*g*h+0.5*m*v^2=0.5*k*d^2-m*g*d

==>g*h=0.5*k*d^2-g*d-0.5*v^2

==>h=(0.5*k*d^2-g*d-0.5*v^2)/g

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