A block of mass 5kg is dragged along a level rough surface having coefficient of friction by a hanging block mass 10kg as
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block of mass 10 kg is placed on a horizontal surface is connected by a cord passing over a frictionless pulley to a hanging block of mass 10 kg. If the coefficient of friction between the block and the surface is 0.5 and g = 10 m/s^2, then find the accleration of the system.
Asked by acv27joy | 22nd May, 2018, 07:35: PM
Expert Answer:
For mass A: T- μm1g = m1a
→ T =m1a + μm1g...(1)
For mass B: m2 g - T = m2a
→ T =m2 g - m2a...(2)
From (1) and (2)
m1a + μm1g=m2 g - m2a
a equals space fraction numerator left parenthesis m subscript 2 minus mu m subscript 1 right parenthesis g space over denominator m subscript 1 plus m subscript 2 end fraction
a equals space fraction numerator left parenthesis 10 minus 0.5 cross times 10 right parenthesis 10 over denominator 10 space plus space 10 end fraction
T h u s comma space space a space equals space 2.5 space m divided by s squared
Thus, the acceleration of the system is 2.5 m/s2
Answered by Shiwani Sawant | 23rd May, 2018, 04:06: PM
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