Question

Plzzz solve the problem...... ​

Answer 1

QUESTION:-

• Two blocks of mass 1 kg each are joined by a string and placed on two inclined planes. The string passess over a pulley. Co efficient of friction between the blocks and the two planes is 0.1. Find (i) acceleration of the system and (ii) tension in the string.

ANSWER:-

• (i) Acceleration of the system = 0.244 m/s².

• (ii) Tension in the string = 6 N.

GIVEN:-

• Two blocks of mass 1 kg each are joined by a string and placed on two inclined planes.

• The string passess over a pulley.

• Co efficient of friction between the blocks and the two planes is 0.1.

EXPLANATION:-

• By free body diagram,
• mg sin θ - T - μ mg cos θ = ma

m = 1 kg

μ = 0.1

θ = 45°

g sin 45° - T - 0.1 g cos 45° = a

T = g sin 45° - 0.1 g cos 45° - a

By free body diagram,

T - mg sin θ - μ mg cos θ = ma

m = 1 kg

μ = 0.1

θ = 30°

T - g sin 30° - 0.1 g cos 30° = a

T = g sin 30° + 0.1 g cos 30° + a

Equate both the T values:-

g sin 30° + 0.1 g cos 30° + a = g sin 45° - 0.1 g cos 45° - a

2a = g sin 45° - 0.1 g cos 45° - g sin 30° - 0.1 g cos 30°

2a = g(sin 45° - 0.1(cos 45°) - sin 30° - 0.1 cos 30°)

g = 9.8 m/s²

2a = 9.8(sin 45° - 0.1(cos 45°) - sin 30° - 0.1 cos 30°)

a = 4.9(sin 45° - 0.1(cos 45°) - sin 30° - 0.1 cos 30°)

sin 45° = cos 45° = 1/√2 = 0.707

sin 30° = 1/2 = 0.5

cos 30° = √3/2 = 1.732/2 = 0.866

a = 4.9(0.707 - 0.1(0.707) - 0.5 - 0.1(0.866))

a = 4.9(0.707 - 0.0707 - 0.5 - 0.0866)

a = 4.9(0.707 - 0.6573)

a = 4.9(0.0497)

a = 0.2435 ≈ 0.244 m/s²

T = g sin 30° + 0.1 g cos 30° + a

T = 9.8(1/2) + 0.1(9.8) √3/2 + 0.244

T = 4.9 + 9.8(0.0866) + 0.244

T = 4.9 + 0.848 + 0.244

T = 5.99 ≈ 6 N

• Hence tension in the string = 6 N and acceleration of the system = 0.244 m/s².

Answer 2

Answer:

ANSWER:-

(i) Acceleration of the system = 0.244 m/s².

(ii) Tension in the string = 6 N.

EXPLANATION:-

By free body diagram,

mg sin θ - T - μ mg cos θ = ma

m = 1 kg

μ = 0.1

θ = 45°

g sin 45° - T - 0.1 g cos 45° = a

T = g sin 45° - 0.1 g cos 45° - a

By free body diagram,

T - mg sin θ - μ mg cos θ = ma

m = 1 kg

μ = 0.1

θ = 30°

T - g sin 30° - 0.1 g cos 30° = a

T = g sin 30° + 0.1 g cos 30° + a

Equate both the T values:-

g sin 30° + 0.1 g cos 30° + a = g sin 45° - 0.1 g cos 45° - a

2a = g sin 45° - 0.1 g cos 45° - g sin 30° - 0.1 g cos 30°

2a = g(sin 45° - 0.1(cos 45°) - sin 30° - 0.1 cos 30°)

g = 9.8 m/s²

2a = 9.8(sin 45° - 0.1(cos 45°) - sin 30° - 0.1 cos 30°)

a = 4.9(sin 45° - 0.1(cos 45°) - sin 30° - 0.1 cos 30°)

sin 45° = cos 45° = 1/√2 = 0.707

sin 30° = 1/2 = 0.5

cos 30° = √3/2 = 1.732/2 = 0.866

a = 4.9(0.707 - 0.1(0.707) - 0.5 - 0.1(0.866))

a = 4.9(0.707 - 0.0707 - 0.5 - 0.0866)

a = 4.9(0.707 - 0.6573)

a = 4.9(0.0497)

a = 0.2435 ≈ 0.244 m/s²

T = g sin 30° + 0.1 g cos 30° + a

T = 9.8(1/2) + 0.1(9.8) √3/2 + 0.244

T = 4.9 + 9.8(0.0866) + 0.244

T = 4.9 + 0.848 + 0.244

T = 5.99 ≈ 6 N

• Hence tension in the string = 6 N and acceleration of the system = 0.244 m/s².