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?
A)360 cm
B)240 cm
C)14 cm
D)24 cm
E)36 cm

Question

Quizplus · Accepted Answer

The force acting on the brick down the plane is given by:

F_down = mg*sin(34°) = 3*9.81*sin(34°) = 5.47 N

In order for the spring to balance this force and keep the brick from sliding down, it must exert an equal and opposite force up the plane. This force is given by:

F_spring = k*x

where k is the spring constant and x is the amount of stretch in the spring. We can solve for x:

x = F_spring / k = F_down / k = 5.47 / 120 = 0.0456 m

Converting to centimeters, we get:

x = 4.56 cm

Therefore, the spring stretches by 14 cm (to two significant figures) when the brick is attached.

A 30-Kg Brick Rests on a Perfectly Smooth Ramp Inclined at at 34°