Question

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Answer 1

Answer:

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Answer 2

QUESTION:

Two blocks, connected by a weightless string, are sliding down a plane of inclination 37º. The respective masses of the blocks are, m₁ = 4 kg, and m₂ = 2 kg. Coefficient of friction of the 1st block with the plane is 0.75 that of the 2nd block is 0.25. Assuming that the string is taut, find (i) the common acceleration of both the blocks and (ii) tension in the string. Given, sin 37° = 0.6; cos 37° = 0.8.

ANSWER:

• The common acceleration of both the blocks = 1.31 m/s².

• Tension in the string = 5.2 N.

GIVEN:

• Two blocks, connected by a weightless string, are sliding down a plane of inclination 37º.

• The respective masses of the blocks are, m₁ = 4 kg, and m₂ = 2 kg.

• Coefficient of friction of the 1st block with the plane is 0.75 that of the 2nd block is 0.25.

• The string is taut.

• sin 37° = 0.6, cos 37° = 0.8.

TO FIND:

• The common acceleration of both the blocks.

• Tension in the string.

EXPLANATION:

From free body diagram,

m₂g sin θ - T - μ₂R₂ = m₂a

m₂ = 2 kg

μ₂ = 0.25

R₂ = m₂g cos θ

θ = 37°

2g sin 37° - T - 0.25(m₂g cos θ) = 2a

2g sin 37° - T - 0.25(2g cos 37°) = 2a

sin 37° = 0.6

cos 37° = 0.8

2g(0.6) - T - 0.25(2g(0.8)) = 2a

1.2g - T - 0.25(1.6g) = 2a

1.2g - T - 0.4g = 2a

0.8g - T = 2a

T = 0.8g - 2a

From free body diagram,

m₁g sin θ + T - μ₁R₁ = m₁a

m₁ = 4 kg

μ₁ = 0.75

R₁ = m₁g cos θ

θ = 37°

4g sin 37° + T - 0.75(m₁g cos θ) = 4a

4g sin 37° + T - 0.75(4g cos 37°) = 4a

sin 37° = 0.6

cos 37° = 0.8

4g(0.6) + T - 0.25(4g(0.8)) = 4a

2.4g + T - 0.75(3.2g) = 4a

2.4g + T - 2.4g = 4a

T = 4a

Equate the values of T

4a = 0.8g - 2a

6a = 0.8g

g = 9.8 m/s²

6a = 7.84

a = 1.306

a ≈ 1.31 m/s²

Hence the common acceleration of both the blocks = 1.31 m/s².

We know that T = 4a

T = 4(1.31)

T = 5.24 ≈ 5.2 N

Hence tension in the string = 5.2 N.