A mailman performed 296 J of work lifting a box of 1.6 m. How much force did the mailman use?
1 answer:
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Look at the picture for the answer
Answer:
Explanation:
Let T be the tension .
Applying newton's second law on the downward movement of the bucket
mg - T = ma
On the drum , a torque of TR will be acting which will create an angular acceleration of α in it . If I be the moment of inertia of the drum
TR = Iα
TR = Ia/ R
T = Ia/ R²
Replacing this value of T in the other equation
mg - T = ma
mg - Ia/ R² = ma
mg = Ia/ R² +ma
a ( I/ R² +m)= mg
a = mg / ( I/ R² +m)
mg - T = ma
mg - ma = T
mg - m x mg / ( I/ R² +m) = T
mg - m²g / ( I/ R² +m ) = T
mg - mg / ( 1 + I / m R² ) = T
b ) T = Ia/ R²
I = TR² / a
c ) Moment of inertia of hollow cylinder
I = 1/2 M ( R² - R² / 4 )
= 3/4 x 1/2 MR²
= 3/8 MR²
I / R² = 3/8 M
a = mg / ( I/ R² +m)
a = mg / ( 3/8 M + m )
T = Ia/ R²
= 3/8 MR² x mg / ( 3/8 M + m ) x 1 /R²
=