CAN YOU PLEASE SHOW THE STEPS THAT LEAD TO THE ANSWER? THE RIGHT ANSWER IS 2.65.

Question

CAN YOU PLEASE SHOW THE STEPS THAT LEAD TO THE ANSWER? THE RIGHT ANSWER IS 2.65. THANKS IN ADVANCE.

A vertically oriented spring with spring constant k has an equilibrium length xo, so one end is xo above the ground. The spring is compressed an amount 1 from its equilibrium length x0. A small dart of mass M is then launched from the spring and reaches a maximum height yi above the ground. Xo=1.5m, x,-0.6m, and yi-6.7m. The spring is now compressed by a new amountx2 0.3m. What height y2 does the dart reach in this case? Answer in meters Answer: 1.56 The correct answer is: 2.65

Explanation / Answer

x0=1.5m, x1= 0.6m, y1=6.7m

Let us first find spring constant k

Use law of conservation of energy

Initial spring KE = Final gravitational energy

½*kx2 = Mgh

For first case,

½*kx2 = Mgh

½*k*x12 = Mg(y1- x0 + x1)

Plugging values x1 = 0.6,

½*k*0.6^2 = M*9.8*(6.7-1.5+0.6)                      

k/M = [2*9.8*(6.7-1.5+0.6)]/0.6^2

k/M = 315.78 N /m^2

For second case, x2 = 0.3 m we can build the eqn as below,

½*k*0.3^2 = M*9.8*(y2 -1.5-x2)

Plugging values,

k/M = [2*9.8*(y2 –1.5-0.3)]/0.3^2

315.78= [2*9.8*(y2 -1.5+ 0.3)]/0.3^2

y2 = 2.65m

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