Question

A 205-g block is pressed against a spring of force constant 1.18 kN/m until the block compresses the spring 10.0 cm. The spring rests at the bottom of a ramp inclined at 60.0° to the horizontal. Using energy considerations, determine how far up the incline the block moves from its initial position before it stops under the following conditions

Answer 1

L = 3.391m

Step-by-step explanation:

a) Given:

Mass of the block, m = 205g=0.205kg

spring constant, k = 1.18 kN/m

Angle made by the ramp = 60°

displacement of the spring, x = 10cm= 0.1m

Now,

The initial potential energy of the spring = final gravitational potential energy of block

mathematically,

`(1)/(2)kx^2 = mgh`

where,

g = acceleration due to gravity

h = height of the block above the ground = LsinΘ

L is the displacement of the block parallel to the ramp

`(1)/(2)kx^2 = mgLsin\theta`

substituting the values in the above equation we get

`(1)/(2)1.18* 1000* 0.1^2 = 0.205* 9.8* Lsin60^o`

or

`5.9 = 1.739* L`

or

L = 3.391m