A mass of 3.0 kg rests on a smooth surface inclined 34° above the horizontal. It is kept from sliding down the plane by a spring attached to a wall. The spring is aligned with the plane and has a spring constant of 120 N/m. How much does the spring stretch?

The spring stretched by x = 13.7 cm

Explanation:

Given data

Mass = 3 kg

k = 120

Angle = 34°

From the free body diagram

Force acting on the box = mg sin

⇒ F = 3 × 9.81 ×

⇒ F = 16.45 N ------- (1)

Since box is attached with the spring so a spring force also acts on the box.

= k x

= 120 -------- (2)

The net force acting on the body is given by

Since acceleration of the box is zero so

Put the values from equation (1) & (2) we get

16.45 = 120

x = 0.137 m

x = 13.7 cm

Therefore the spring stretched by x = 13.7 cm

the answer is d

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i think it would be 5