A 4kg block sitting on a flat table is attached by a string over a cylindrical p
ID: 1410376 • Letter: A
Question
A 4kg block sitting on a flat table is attached by a string over a cylindrical pulley to a 3kg block which is hanging in the air 4 meters above the ground. The coefficient of the kinetic friction is 0.2. The radius of the wheel is 0.08 meters and its mass is 0.07kg (Assume the pulley to be a solid cylinder). I need help on the step-by-step portion. The answers follow after the questions.
a) Find the acceleration (a) of the system. 3.06m/s^2
b) Find the tensions (T1 and T2) in the upper and lower part of the string, respectively. T1=20.098 Newtons, T2=20.205 Newtons.
c) Find the total work done on the wheel by the strings. 0.43 Joules
d) Find the final angular speed of the wheel. 61.89 Radians/sec
Explanation / Answer
mass of the block on the table m 1 = 4 kg
hanging mass m 2 = 3 kg
coefficient of the kinetic friction = 0.2
we know , m 2* g - T = m 2* a ( or ) T = m 2* g - m 2* a ------( 1) and T - m 1 * g = m 1 * a T = m 1 * a+ m 1 * g from above equations
acceleration of the cylinder a =
= [ ( 3 - 0.2 * 4 ) / ( 4+3) ] 9.8
= [ ( 3-0.8 ) / 7 ] 9.8
= ( 2.2 / 7 ) * 9.8
= 0.314 *9.8
= 3.06m / s^2
b)t1=20.635 N
t2=20.524 N
c)4m=r
=4/0.8=5 radians
work done =(t1-t2)=(20.635-20.524)5=0.555 J
d)=sqrt(2)=sqrt(2*5*3.16/0.08)
=19.875 rad/s
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