Question

A 3.0-kg brick rests on a perfectly smooth ramp inclined at 34° above the horizontal. The brick is kept from sliding down the plane by an ideal spring that is aligned with the surface and attached to a wall above the brick. The spring has a spring constant (force constant) of 120 N/m. By how much does the spring stretch with the brick attached?

Answer 1

Answer:

It stretches in 0.137 m

Explanation:

Using the Hook's law:

F = kx

$F=mgsin\theta \\kx=mgsin\theta\\x=\frac{mgsin\theta}{k}$

Where

m = 3 kg

θ = 34°

k = 120 N/m

Replacing:

$x=\frac{3*9.8*sin34}{120} =0.137m$

Answer 2

Answer: 0.137 m

Explanation:

Given

Mass of brick, m = 3 kg

Angle of inclination, Φ = 34°

Force constant of the spring, k = 120 N/m

The force of the brick, F can be gotten using the relation

F = mg

F = 3 * 9.8

F = 29.4 N

Now, the force parallel to the incline, F(p) can be gotten using the formula,

F(p) = F sinΦ, so that

F(p) = 29.4 * sin 34

F(p) = 29.4 * 0.559

F(p) = 16.4 N

The stretch distance then is,

d = F(p) / k * 1 m

d = 16.4 / 120

d = 0.137 m

Thus, the spring stretched by a distance of 0.137 m