In fig. 8-42, a block of mass m = 12 kg is released from rest on a frictionless incline of angle θ = 30°. below the block is a spring that can be compressed 2.0 cm by a force of 270 n. the block momentarily stops when it compresses the spring by 5.5 cm. (a) how far does the block move down the incline from its rest position to this stopping point? (b) what is the speed of the block just as it touches the spring?

a) The block moves down a distance of 35 cm down the incline from its rest position to this stopping point

b)  The speed of the block just as it touches the spring is 1.7 m/s

Explanation:

First let us determine the spring constant

Given: and

⇒

Let the distance the block moves down the incline from its rest position to this stopping point be L measured along the incline

and height of the block be H from the stopping point

a)Given that ° and where x is the distance by which the spring compresses

⇒

and using law of conservation of energy:

Energy at rest point=Energy at stopping point

⇒

⇒

Therefore,

⇒

Therefore the block moves down a distance of 35 cm down the incline from its rest position to this stopping point

b) where is the compression in spring and is the distance between starting of spring to the block

∴

Let h be the height of the block from the starting of spring

∴

using law of conservation of energy:

Energy at rest point=Energy at starting point of the spring

⇒ where is the speed of the block

⇒

gpe = mass x gravity constant x height or… gpe = mgh

where… mass: measured in kilograms (kg)

gravity: gravity constant on earth = 10.00 m/s2

height: measured in meters (m)

gpe: measured in joules (j) or kilojoules (kj)

how do you solve gpe?

gravitational potential energy (gpe) is the energy an object has because of its position above the. ground.

gpe = mass x gravity constant x height or… gpe = mgh.

mass:

note: to earn full marks when solving science word problems, you must show your work:

formula.

example:

given values:

mass = 0.3 kg.