Question

A mass of 3.0 kg rests on a smooth surface inclined 28o 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:

$x = 0.115 m$

Explanation:

As we know that the mass is pulled downwards due to its own weight along the inclined plane

since mass is not sliding downwards so we can say that the weight along the inclined plane is counterbalanced by spring force

so we will have

$mg sin\theta = kx$

now we will have

$3(9.81)(sin28) = kx$

now we have

$x = \frac{3(9.81) sin28}{k}$

here we know that

k = 120 N/m

so we have

$x = \frac{3(9.81) sin28}{120}$

$x = 0.115 m$

Answer 2

Answer:x=11.50 cm

Explanation:

Given

mass of object(m)=3 kg

inclination$=28^{\circ}$

spring constant(k)=120 N/m

Spring Resist the sin component of weight

Thus

$k x=W\sin \theta$

$120\times x=3\times 9.8\times \sin 28$

$x=11.50 cm$