A 1 kg block situated on a rough incline is connected to a spring cons

1.	A 1 kg block situated on a rough incline is connected to a spring constant 100Nm-1.The block is released from rest with the spring in the unstretched position.The block moves 10cm down the incline before coming to rest. Find the coefficient of friction between the block and the incline.
Answer» The normal reaction force of block on incline R = mg cosθ Let the coefficient of friction be M ∴ F = MR = μ mg cosθ ∴ Net force on block = mg sinθ – F = mg (sinθ – μ cosθ) we know that the block moves by distance x = 10cm = 0.1 m. The work done by net force in moving block 0.1m is = Energy stored in the spring. ∴ mg (sin θ – μ cos θ) x = \(\frac{1}{2}\) kx² ∴ 2 mg (sin 37° – μ cos 37°) = kx ⇒ 2 × 1 × 9.8 m^s-2 (sin37°- μ cos37°) = 100 × 0.1m ⇒ 19.6 (0.601 – μ × 0.798) = 10 0.601 – μ × 0.798 = 0.5102 ⇒ μ × 0.798 = 0.09079 ∴ μ = \(\frac{0.09079}{0.798}\)= 0.1137

A 1 kg block situated on a rough incline is connected to a spring constant 100Nm-1.The block is released from rest with the spring in the unstretched position.The block moves 10cm down the incline before coming to rest. Find the coefficient of friction between the block and the incline.

Answer»

The normal reaction force of block on incline R = mg cosθ

Let the coefficient of friction be M

∴ F = MR

= μ mg cosθ

∴ Net force on block = mg sinθ – F

= mg (sinθ – μ cosθ)

we know that the block moves by distance x = 10cm = 0.1 m.

The work done by net force in moving block 0.1m is

= Energy stored in the spring.

∴ mg (sin θ – μ cos θ) x = \(\frac{1}{2}\) kx²

∴ 2 mg (sin 37° – μ cos 37°) = kx

⇒ 2 × 1 × 9.8 m^s-2 (sin37°- μ cos37°) = 100 × 0.1m

⇒ 19.6 (0.601 – μ × 0.798) = 10

0.601 – μ × 0.798 = 0.5102

⇒ μ × 0.798 = 0.09079

∴ μ = \(\frac{0.09079}{0.798}\)= 0.1137

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