Question

9.
motor to elevad!
A block of mass 4 kg is released from the top
an inclined smooth surface as shown in figure
spring constant of spring is 150 N/m and block
comes to rest after compressing spring by im
then find the distance travelled by block before
comes to rest
Oooo3004
30°
Tower A D
D​

Answer 1

Answer:

energy lost by the block=energy gained by the spring

energy gained= 1/2 x 150 x 1^2=75J

energy of the block= mgh

mgh=75J

h=75/(4 x 10)

1.09m

sin 30°=perp/hypo

hypo=(1/2)/1.09

=0.46m

therefore total distance travelled= 0.46 + 1(while compressing the spring)

1.46m(ans)