Student Question

# If spring 2 can only compress by 2 meters, what mass must the block have to compress the spring by exactly 2 meters? Assume the block starts from rest.

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When the body is released from the rest position, the elastic potential energy is transformed into kinetic energy and gravitational potential energy. At the position hi = Δx1, the kinetic energy has the following value:

EK = Ee - Eg

Ee, is the elastic potential energy in the initial position.

Eg, is the gravitational potential energy at the height hi.

So we have the following equation:

m(v^2)/2 = K1(hi^2)/2 – mghi        (1)

v, is the velocity of the body.

g, is the acceleration of gravity.

m, is the mass of the body.

When the body has compressed the spring K2, all the kinetic energy has been turned into elastic potential energy and gravitational potential energy. For this position, we can consider the following equation:

EK = Ee + Eg

m(v^2)/2 = K2(Δx^2)/2 + mg(Δx + 41)       (2)

With the equations 1 and 2 we can establish the following equality:

K1(hi^2)/2 – mghi= K2(Δx^2)/2 + mg(Δx + 41)

K1(hi^2)/2 - K2(Δx^2)/2 = mg(Δx + 41) + mghi

Substituting the values:

(71)(25)/2 – (37)(4)/2 = m(9.8)(43) + m(9.8)(5)

813.5 = m(470)

m = 1.73 kg