A little mass and a big Mass are attached by a string on an incline surface as p

Question

A little mass and a big Mass are attached by a string on an incline surface as picture shown below. The system is static initially.If m= 5kg and M =10kg. The angle of the incline is 30 degree.(g=10m/s2).(you do not need to consider about the positive and negative sigh or units or friction during calculation)

a. How much is the net force on little mass? (in N)

b. What is the acceleration of the little mass? (in m/s^2)

c. If big Mass moves 2.5 meters downward, what is the velocity of the little mass? (in m/s)

d. How long does it take for big Mass to move 2.5 meters? (in s)

e. If the angle of the incline varies, is the relationship between acceleration vs the angle of the incline linear or nonlinear? if linear, please mark as L. if nonlinear please mark as N.

Explanation / Answer

Given
  
   little mass m = 5 kg, bigger mass M = 10 kg

   the angle of inclination is theta = 30 degrees

acceleration due to gravity is g = 10 m/s2

from the given free body diagram

   for the little mass

   T - mg sin theta= ma

for bigger mass

   Mg -T = Ma

both having same acceleration so

T - mg sin theta = Mg -T

2T = g(M+m sin theta)

T = 0.5*g(M+m sin theta)

T = 0.5*10(10+5sin30) N

T = 62.5 N

substitute T in Mg -T = Ma ==> a = (10*10-62.6)/10 m/s2 = 3.74 m/s2

a) the net force on the little mass is

   F = T- mg sin theta

   F = 62.5 - 5*10sin30 N = 37.5 N
b) acceleration of little mass a = 3.74 m/s2

c) if s = 2.5 m downward , the velocity of little mass is

   v = sqrt(2as) = sqrt(2*3.74*2.5) m/s = 4.324 m/s
d)

   time taken for the big mass to reach 2.5 m downward is

   s = ut + 0.5at^2

   2.5 = 0*t+ 0.5*3.74*t^2

   t = 1.156 s

e) if the angle of inclination changes the relationship between acceleration vs the angle of the incline linear

L

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