Aspring (k = 600 n/m) is at the bottom of a frictionless plane that makes an angle of 30° with the horizontal. the upper end of the spring is depressed 0.10 m, and a 2.0-kg block is placed against the depressed spring. the system is then released from rest. what is the kinetic energy of the block at the instant it has traveled 0.10 m and the spring has returned to its uncompressed length?

kinetic energy  is 2.02 J

Explanation:

Given data

k = 600 N/m

angle = 30°

spring = 0.10 m

mass = 2 kg

to find out

kinetic energy of the block

solution

we know initial energy that is

Energy 1 = 1/2 × kx²

put here k = 600 and x = 0.1

energy 1 = 1/2 × 600(0.1)²

energy  = 3

potential energy  PE = mgh

PE = 2 ×9.8 ×0.1sin30

PE = 0.98 J

so we know final energy is

E = KE of block + PE

so

kinetic energy  = 3 - 0.98

kinetic energy  = 2.02 J

where's the following? you gotta add it to know what the choices are

explanation:

increasing transnational kinetic energy. because the transnational kinetic energy formula consists of only two variables, mass and velocity, increasing one of those properties is the only way to increase an object's transnational kinetic energy. increases to mass and velocity, however, do not have the same impact.