The lower block in the figure below is pulled on by a rope with a tension force
ID: 2054281 • Letter: T
Question
The lower block in the figure below is pulled on by a rope with a tension force of 39 N. The coefficient of kinetic friction between the lower block and the surface is 0.22. The coefficient of kinetic friction between the lower block and the upper block is also 0.22. What is the acceleration of the 2.0 kg block?
----- m/s2?
Explanation / Answer
Equation for block 1 is
F = M*A
The force for block 1 is
Tension - (friction coefficent*mass*g)
the mass is 1 kg
the a is unknown, but we know that both blocks have the same acceleration, since they are connected.
So here is the final equation that describes the acceleration of block 1:
T-.22(1kg*9.8) =1kg*a
T-.22(9.8) = a---------(1)
now lets find an equation that describes the acceleration of block 2:
F = ma
39N - T -(.22*3kg*9.8 + .22(1kg*9.8) = 2kg*a
so:
a = {39N - T -(.22*3kg*9.8 + .22(1kg*9.8) } / 2
using equation (1)
2*T-.22*2*(9.8) ={39N - T -(.22*3kg*9.8 + .22(1kg*9.8) }
2T-4.132=39-T-8.642
3T=34.508
T=11.563
T - (.22*g) = a
11.563 -(2.156) = a
a =9.407
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.