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Open Box Problem Courseworks

Assignment 7Answer: F = p/t = m(vf– vi) / t = 17.5(0 – 3.50)/8.76 = -7.0 NProblem 9: Two blocks of masses m and 3m are placed on a frictionless, horizontal surface. A light spring is attached to the more massive block, and the blocks are pushed together with the spring between them (Fig. P9.9). A cord initially holding the blocks together is burned; after that happens, the block of mass 3m moves to the right with a speed of 2.00 m/s. (a) What is the velocity of the block of mass m? (b) Find the system’s original elastic potential energy, taking m 5 0.350 kg. (c) Is the original energy in the spring or in the cord? (d) Explain your answer to part (c). (e) Is the momentum of the system conserved in the bursting-apart process? Explain how that is possible considering (f) there are large forces acting and (g) there is no motion beforehand and plenty of motion afterward?Answer:a) Momentum is conserved and we have: mvm+ 3mV3m= 0vm= (− 3m/m) * V3m= −3V3m= −3 × 2.0 i = −6 i|vm| = 6.0 m/s b) Total energy is conserved, gravitation potential energy doesnt change but elastic potentialenergy changes after the spring is released. Uelf − Ueli = −(Kf − Ki) 0 − Uelf = −( ½*mvm2+ ½* 3mV3m2) Ueli = m/2 (vm2+ 3V3m2) = 0.350*(62+ 3 × 22) = 16.8J c) the original energy is in the spring.

Problem 1: A sheet of metal 12 inches by 10 inches is to be used to make a open box. Squares of equal sides x are cut out of each corner then the sides are folded to make the box. Find the value of x that makes the volume maximum.

Solution to Problem 1:

  • We first use the formula of the volume of a rectangular box.

    V = L * W * H

  • The box to be made has the following dimensions:

    L = 12 - x

    W = 10 - 2x

    H = x

  • We now write the volume of the box to ba made as follows:

    V(x) = x (12 - 2x) (10 - 2x) = 4x (6 - x) (5 - x)

    = 4x (x 2 -11 x + 30)

  • We now determine the domain of function V(x). All dimemsions of the box must be positive or zero, hence the conditions

    x > = 0 and 6 - x > = 0 and 5 - x > = 0

  • Solve the above inequalities and find the intersection, hence the domain of function V(x)

    0 < = x < = 5

  • Let us now find the first derivative of V(x) using its last expression.

    dV / dx = 4 [ (x 2 -11 x + 3) + x (2x - 11) ]

    = 3 x 2 -22 x + 30

  • Let us now find all values of x that makes dV / dx = 0 by solving the quadratic equation

    3 x 2 -22 x + 30 = 0

  • Two values make dV / dx = 0: x = 5.52 and x = 1.81, rounded to one decimal place. x = 5.52 is outside the domain and is therefore rejected. Let us now examine the values of V(x) at x = 1.81 and the endpoints of the domain.

    V(0) = 0 , v(5) = 0 and V(1.81) = 96.77 (rounded to two decimal places)

  • So V(x) is maximum for x = 1.81 inches. The graph of function V(x) is shown below and we can clearly see that there is a maximum very close to 1.8.



More references on calculus problems

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