Respuesta :
Answer: The distance along the incline the mass slide from the point of release until it is brought momentarily to rest is 0.68 meter
Explanation:
Let distance slide by block along the incline be L
[tex]\therefore h=L\sin (37^{\circ})[/tex]
where h = initial height of block from the ground level
Since the inclined surface is frictionless so total mechanical energy is conserved for the spring-block system
Therefore applying conservation of mechanical energy we get
[tex]PE_{si}+PE_{gi}+KE_i=PE_{sf}+PE_{gf}+KE_f[/tex]
where
[tex]PE_{si}[/tex] = Initial spring potential energy
[tex]PE_{gi}[/tex] = Initial gravitational potential energy
[tex]KE_i[/tex] = Initial kinetic energy
[tex]PE_{sf}[/tex] = Final spring potential energy
[tex]PE_{gf}[/tex] =Final gravitational potential energy
[tex]KE_f[/tex] = Final kinetic energy
=>[tex]0+mgL\sin (37^{\circ})+0=\frac{kx^{2}}{2}+0+0[/tex]
=>[tex]1.0\times 9.8\times L\sin (37^{\circ})=\frac{200\times 0.20^{2}}{2}[/tex]
[tex]\therefore L= 0.68 m[/tex]
Thus the distance along the incline the mass slide from the point of release until it is brought momentarily to rest is 0.68 meter
