A 7kb kg (m1) block rests on an inclined plane. The plane makes an angle of 20 d
ID: 581462 • Letter: A
Question
A 7kb kg (m1) block rests on an inclined plane. The plane makes an angle of 20 degrees with the horizontal. The 7 kg block is tied to a second block (m2= 2 kg) which hangs over the end of the inclined plane after the rope passes over an ideal pulley.
a) Determine the magnitude of the accerleration of the blocks assuming there is no friction.
b) Same figure but a=0.134m/s2. Find the magnitude of kinetic friction force between the second block and the ledge.
What is the magnitude of coefficient of kinetic friction between the block and the ledge?
Explanation / Answer
a) the component of weight of m1 parallel to surface responsible for tension in string = mg sin theta
= 7 * 9.81 * sin 20 = 23.44
Acceleration of blocks
a = ( m1 sin 20 - m2 ) g /( m1 sin20 + m2 )
= (7sin20 - 2 ) 9.81 / ( 7 sin20 + 2 )
= 0.88 ms^-2
b) frictional focre on block 1 is responsible for the decrease in acceleration
Decrease in acceleration = 0.88 - 0.134 = 0.747
F = m × a
Kinetic friction = 7 sin 20 × 0.747
= 1.78 N
c ) coefficient of kinetic friction = kinetic friction / normal reaction
= 1.78 / ( mg cos 20 )
= 1.78 / 7 * 9.81 * cos 20
=0.028
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