Physics

The figure shows a 7 kg block being pulled along a frictionless floor by a cord that applies a force of constant magnitude 20 N but with an angle θ(t) that varies with time. When angle θ = 29°, at what rate is the block's acceleration changing if (a)θ(t) = (3 × 10-2 deg/s)t and (b)θ(t) = (-3 × 10-2 deg/s)t ? (Hint: Switch to radians.)

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asked by John
  1. I assume that Theta is up and down from horizontal
    calling theta = A
    d sin A /dA = cos A
    d sin A/dt = d sin A/dA * dA/dt
    so
    d sin A /dt = cos A (dA/dt )
    similarly
    d cos A/dt = -sin A dA/dt
    That settled, let's look at the problem
    Force = m * a
    a = F/7
    da/dt = (1/7) dF/dt
    F in direction of motion = 20 cos A
    d F/dt =20 d/dt(cosA)
    but we already know what that is
    dF/dt = 20 (-sin A) dA/dt
    A = 28 deg = .489 radian
    sin A = .469
    dA/dt = 3*10^-2 (pi/180) = 5.24*10^-5 rads/s
    so
    dF/dt = 20 (-.489)(5.24*10^-5)
    da/dt =(1/7)dF/dt = (20/7)(-.489)(5.24*10^-5)

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    posted by Damon
  2. Net force=mass*acceleration
    20*cosTheta=mass* acceleratin
    take the deriviative...

    -20SinTheta dTheta/dt=mass*d(acceleration)/dt

    yes, dTheta/dt is in radians/sec, so convert it. solve for d(acceleration)/dt and the units will be m/sec^3

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