Answer:
(a) a = 1.875 m/s²
(b)T = 1.575 N
(c)T increase
Explanation:
Newton's second law:
∑F = m*a Formula (1)
∑F : algebraic sum of the forces in Newton (N)
m : mass s (kg)
a : acceleration (m/s²)
We define the x-axis in the direction parallel to the movement of the blocks on the horizontal surface and the y-axis in the direction perpendicular to it.
Forces acting on the block A
WA: Weight of of the A block : In vertical direction downaward (-y)
NA : Normal force of the A block :In vertical direction upaward (+y)
F= 5.7 N In in the direction parallel to the movement of the blocks (+x)
T : tension of the string: In in the direction (-x)
Forces acting on the block B
WB: Weight of the B block: In vertical direction downaward (-y)
NB : Normal force of the B block :In vertical direction upaward (+y)
T : tension of the string: In in the direction (+x)
Data
mA = 2.2 kg
mB = 0.84 kg
(a) Magnitude of the acceleration of the blocks
Newton's second law to B block:
∑Fx = m*a
T = (0.84)*a Equation (1)
Newton's second law to A block:
∑Fx = m*a
5.7 - T = (2.2)*a We replace T of the Equation (1)
5.7 - (0.84)*a = (2.2)*a
5.7 = (2.2)*a + (0.84)*a Equation (2)
5.7 =(3.04)*a
a =5.7 / (3.04)
a = 1.875 m/s²
(b)Tension (in N) in the string connecting the two blocks
We replace data in the Equation (1)
T = (0.84)*a
T = (0.84)*(1.875)
T = 1.575 N
(c) How will the tension in the string be affected if mA is decreased?
We observed Equation (2)
5.7 = (2.2)*a + (0.84)*a
5.7 = (mA)*a + (0.84)*a
5.7 =a*( mA+ 0.84)
if mA decrease , then, the acceleration increase and T increase
