Two blocks A and B with mA = 2.2 kg and mB = 0.84 kg are connected by a string of negligible mass. They rest on a frictionless horizontal surface. You pull on block A with a horizontal force of 5.7 N. (a) Find the magnitude of the acceleration (in m/s2) of the blocks. m/s2 (b) Determine the tension (in N) in the string connecting the two blocks. N (c) How will the tension in the string be affected if mA is decreased? increase decrease remain the same

Question

Carissa

Physics

5 October, 18:05

Two blocks A and B with mA = 2.2 kg and mB = 0.84 kg are connected by a string of negligible mass. They rest on a frictionless horizontal surface. You pull on block A with a horizontal force of 5.7 N. (a) Find the magnitude of the acceleration (in m/s2) of the blocks. m/s2 (b) Determine the tension (in N) in the string connecting the two blocks. N (c) How will the tension in the string be affected if mA is decreased? increase decrease remain the same

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Answers (1)

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Everett Hanna · Answer 1

(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