A block of mass m_1m1 = 7.6 kg, sliding down a ramp inclined at angle \\theta =

Question

A block of mass m_1m1 = 7.6 kg, sliding down a ramp inclined at angle heta = 32 degrees, is connected by a light string over a pulley to a second mass, m_2 m2 = 1.5 kg, suspended along the vertical side of the incline. There is a (kinetic) frictional force on the first mass: F_{fr, ~ ramp ~ on ~ 1} = 0.2 cdot F_{N, ~ ramp ~ on ~ 1} (The frictional force is proportional to the normal force.) Calculate the acceleration of the first mass, m_1m1. Your answer should be positive if the acceleration is up the ramp, and negative if the acceleration is down the ramp. Give your answer in the units of m/s2 to at least three significant digits. You won't be graded on the number of digits, this is just so your answer is not counted as incorrect due to rounding. Note: Initially the mass m_1m1 is sliding down the ramp and the mass m_2m2 is moving upwards.

Explanation / Answer

Given

m1 = 7.6 kg, m2 = 1.5 kg

coefficient of kinetic friction i s mue_k = 0.2

the angle theta = 32 degrees

the equation of force on each mass as the m1 sliding down the ramp i s

m1

   -m1 g sin theta + T+mue_k*m1g cos theta = m1a

and

m2
   -m2g +T = - m2a

   T = m2(g-a)

substitute in m1 eq

   -m1 g sin theta + m2(g-a)+mue_k*m1g cos theta = m1a

   -m1 g sin theta + m2g-m2a+mue_k*m1g cos theta = m1a
  
   -a(m1+m2)-g(m1 sin theta -m2 - mue_k*m1 cos thet) =0

   a(m1+m2) = - g(m1 sin theta -m2 - mue_k*m1 cos thet)

   a = - g(m1 sin theta -m2 - mue_k*m1 cos thet)/(m1+m2)

substituting the values

   a = -9.8(7.6sin32-1.5 -0.2*7.6cos32)/(7.6+1.5) m/s2

   a = -1.333611 m/s2

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