In the figure below the two blocks are connected by a string of negligible mass
ID: 1481114 • Letter: I
Question
In the figure below the two blocks are connected by a string of negligible mass passing over a frictionless pulley. m1 = 10.0 kg and m2 = 5.60 kg and the angle of the incline is = 40.0°.Assume that the incline is smooth. (Assume the +x direction is down the incline of the plane.)
(a) With what acceleration does the mass m2 move on the incline surface? Indicate the direction with the sign of your answer.
_________ m/s2
(b) What is the tension in the string?
__________N
(c) For what value of m1 will the system be in equilibrium?
___________kg
Explanation / Answer
a) Let T is the tension in the string.
net force acting on m1, Fnet1 = m1*g - T
m1*a = m1*g - T
==> T = m1*g - m1*a ---(1)
Net force acting on m2, Fnet2 = T - m2*g*sin(40)
m2*a = T - m2*g*sin(40)
T = m2*a + m2*g*sin(40) ---(2)
from equatiosn1 and 2
m1*g - m1*a = m2*a + m2*g*sin(40)
a*(m1 + m2) = m1*g - m2*g*sin(40)
a = (m1*g - m2*g*sin(40))/(m1 + m2)
= (10*9.8 - 5.6*9.8*sin(40))/(10 + 5.6)
= 4.02 m/s^2 <<<<<<<<<<-----------------------Answer
b) from equation 1
T = 10*9.8 - 10*4.02
= 57.8 N <<<<<<<<<<-----------------------Answer
c) when the system is in equilibrium, a = 0
(m1*g - m2*g*sin(40))/(m1 + m2) = 0
m1*g - m2*g*sin(40) = 0
m1 = m2*sin(40)
= 5.6*sin(40)
= 3.6 kg <<<<<<<<<<-----------------------Answer