IMPOSSIBLE PHYSICS PROBLEM!?

3 Answers

• SoCalGuy

  10 hours ago

  Before doing any math, let’s look at the problem from this perspective: We need to find how fast it takes package A to reach the bottom. Package A will accelerate down the ramp along with Package B. BOTH Package A AND Package B MUST have the same acceleration down the ramp. Since Pacakge B is bigger and has less kinetric friction (0.170 compared to 0.220), it’s going to push Package A down the ramp. Both packages are thus going down the ramp at the same speed. By knowing the acceleration of both packages, we can find how long it’ll take Package A to reach the bottom (if we were told to find the time it took Package B to travel down, it would take longer than the given 2 m since Package B is behind Package A).

  – – – – – – – – – – – – – – – – – – – — – – — – — – – – — – – – – – – – – — – – – –

  PACKAGE A:

  First, draw a free body diagram of Package A. If drawn correctly, our equations are :

  ΣFx = max (the sum of the forces in the direction of motion equals the product of mass and acceleration)

  (1) m1gsin(20°) + Fb – Ff_A = m1a

  where:

  m1 = mass of Package A (6 kg)

  Fb = the force of Package B on Package A (remember, Package B is pushing Package A so a pushing force is present)

  Ff_A = the frictional force on Package A

  a = acceleration

  ΣFy = 0

  (2) N – m1gcos(20°) = 0

  However, Ff_A = μk1*N = μk1*m1gcos(20°)

  where: μk1 = coefficient of kinetic friction on Package A (0.220). Substitute into eq. (1):

  (3) m1gsin(20°) + Fb – μk1*m1gcos(20°) = m1a

  – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –

  PACKAGE B:

  The same procedure applies to Package B:

  ΣFx = max

  m2gsin(20°) – Fa – Ff_B = m2a

  where: Fa = Force of Package A on Package B (an opposing reaction force since Package B applies a force on Package A)

  (4) m2gsin(20°) – Fa – μk2*m2gcos(20°) = m2a

  – – – – – – – – – – – – – – – – – – – — — — – — – – — – – – – – – – – –

  SOLVE:

  (3) m1gsin(20°) + Fb – μk1*m1gcos(20°) = m1a

  (4) m2gsin(20°) – Fa – μk2*m2gcos(20°) = m2a

  Add equations (3) and (4) together. Note that Fb = Fa. These are two opposing reaction forces that MUST be equal for both boxes to slide down together.

  m1gsin(20°) + m2gsin(20°) – μk1*m1gcos(20°) – μk2*m2gcos(20°) = m1a + m2a

  gsin(20°)(m1 + m2) + gcos(20°)(-μk1*m1 – μk2*m2) = a (m1 + m2)

  a = [gsin(20°)(m1 + m2) + gcos(20°)(-μk1*m1 – μk2*m2)] / (m1 + m2)

  a = [9.81*sin(20°)(6+8) + 9.81*cos(20°)(-0.220*6 – 0.170*8)] / (6 + 8)

  a = 1.59 m/s^2

  This is the acceleration of both blocks. Now, we can find the time it takes package A to reach the bottom:

  x = vix*t + 0.5*ax*t^2

  where:

  x = distance (2 m)

  vix = initial velocity of Package A (0 m/s,assumed to start from rest)

  ax = 1.59 m/s^2

  t = time it takes for the package to reach the bottom

  2 = 0.5*1.59*t^2

  ==>t = 1.586 seconds

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• PQR Theorist

  10 hours ago

  I made it 1.59 seconds.

  Resolve forces normal and parallel to the slope.

  normal force is 6 g cos(20°) for package A and 8 g cos(20°) for package B

  frictional forces on the system are therefore (6*.22+8*.17) g cos(20°)=2.518g (acting upslope)

  the gravity component down the slope is 14 g sin(20°)=4.788 g

  Net downslope force=4.788g-2.518g=2.27g

  Mass of system=14, so acceleration f =2.27g/14=1.59 m/s² downslope

  (NOTE: This wouldn’t work if package A was more slippery than B, as then they would separate going down the slope)

  distance travelled = 2=1/2 f t² so t=1.59 seconds approx.

  2
• severino

  4 days ago

  At consistent velocity, the internet stress = 0 internet stress = 0 = utilized stress – Friction stress. (F) utilized stress = Friction stress. F = mass * g * u(ok) F = 8 (kg) * 9.8 (m/sec^2) * 0.14 = 10.ninety 8 (kg*m/sec^2) or 10.ninety 8 Newtons.

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