Homework Assignment 6 (due Friday 2/28/14) These problems deal potential energy,

Question

Homework Assignment 6 (due Friday 2/28/14)

These problems deal potential energy, force and rockets

Problem 1. Find the increase in potential energy when the mass moves from the bottom of the incline to the top of the incline.

Problem 2. Mass M is at position x1 and the equilibrium position of the spring is L0. Find the change in potential energy when the mass moves from x1 to x2.

Problem 3. A force is given by: where A and B are two constants

(A) Prove whether or not this force is conservative.

(B) Find the work done moving a mass from (0,1) to (1,1) along segment 1 and then from (1,1) to (1,2) along segment 2. The coordinates are in meters.

Problem 4. Rocket problem [requires calculus]

As shown by Mr. T. Sanders in class, the equation of motion of a rocket in an inertial frame is:

This equation applies to a rocket in a force free region of space. In this problem, assume that the rocket loses mass at a constant rate and therefore the rocket's mass at any time is: where k is a constant.

(A) Insert this last equation into the equation of motion and obtain a simple first order differential equation for the rocket's velocity. Ans:

(B) 'Separate' this equation and integrate to find the rocket's velocity as a function of time. Evaluate the constant of integration by realizing that at time t = 0, u = 0 (the rocket starts from rest) and the mass at t = 0 is .

Explanation / Answer

3)

Force is conservative , If curl of force is zero ,   or work done by force in closed path is zero   or work done by force only depend on initial and final position ,

all above statement is same , for conservative force .

F = Ai + Bj

    curl of F   = ( d/dx i + d/dy j) x F

                     = ( d/dx i + d/dy j) x ( Ai + Bj)

                    =0

hence F is conservative

B) If   F = force = A i + B j

W1 =work done along (0,1) to (1,1)    = integration [ F*dr ]

     = integration [ ( Ai + Bj)*( dx i + dy j) ]

     = integration [   Adx] + integration[Bdy]   , where x =0 to 1 , y =1 to 1

     = AX + BY   , where x =0 to 1 , y =1 to 1

   = A(1 -0 ) + B(1-1)

= A

W2 =work done along (1,1) to (1,2)     = integration [ F*dr ]

     = integration [ ( Ai + Bj)*( dx i + dy j) ]

     = integration [   Adx] + integration[Bdy]   , where x =1 to 1 , y =1 to 2

      = A(1 -1 ) + B(2-1)

     = B

4)   Mo = initial mass of rocket

       k = - dM/dt = rate of change of mass

       V =gas velocity ,

     du/dt = change in velocity of rocket

               = kV/( Mo-kt)

B)    du = kV/( Mo-kt) *dt

       integrate on both side ::

        integration [du] =integration[ kV/( Mo-kt) *dt ]

                 U = kV *integration[ 1/( Mo-kt) *dt ]

                  U = kV ln ( Mo-kt) + C     , where C = constant of integration

apply initial condition , at t =0 , u =0 , THEN

                    0 = kV ln ( Mo) + C

                   C   = - kV ln ( Mo)

so    U = kV ln ( Mo-kt) - kV ln ( Mo)

= kV ln[ ( Mo-kt) /Mo]

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