fernandez122115
31.03.2020 •
Physics
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?
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Ответ:
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
Ответ: