Question

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?

Answer 1

Answer:

The spring stretched by x = 13.7 cm

Explanation:

Given data

Mass = 3 kg

k = 120 $\frac{N}{m}$

Angle $\theta$ = 34°

From the free body diagram

Force acting on the box = mg sin$\theta$

⇒ F = 3 × 9.81 × $\sin34$

⇒ F = 16.45 N ------- (1)

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

$F_{sp}$ = k x

$F_{sp}$ = 120 $x$ -------- (2)

The net force acting on the body is given by

$F_{net} = ma$

Since acceleration of the box is zero so

$F_{net} = 0$

$F - F_{sp} = 0$

$F = F_{sp}$

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

16.45 = 120$x$

x = 0.137 m

x = 13.7 cm

Therefore the spring stretched by x = 13.7 cm

Answer 2

The spring stretched by "13.7 cm". To understand the calculation, check below.

Force and spring constant

According to the question,

Mass, m = 3 kg

Spring constant, k = 120 N/m

Angle, θ = 34°

We know the relation,

Force acting on box be:

= mgSinθ

By substituting the values,

= 3 × 9.81 × Sin34°

= 16.45 N

The spring force be:

→ $F_{sp}$ = kx

        = 120 x

We know,

$F_{net}$ = ma

Since, acceleration is Zero,

F - $F_{sp}$ = 0 or,

       F = $F_{sp}$

By substituting the values,

 16.45 = 120 x

        x = $\frac{120}{16.45}$

           = 13.7 cm

Thus the approach above is appropriate.

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