Question

a block of mass 12 kg is released from rest om a friction-less incline of the angle theta=30
below the block is a spring having spring constant 10^4 N/m. the block stops momentarily when it compresses the spring by 4 cm. how far does the block move down the incline from its rest position to this stopping point

Answer 1

Gravitational potential converts to spring potential as block slides and spring compresses. So balance the 2 taking account of slope.

Answer 2

Potential energy of compressed spiring=.5kx^2=.5*10^4*.04^2=?

Answer 3

Potential energy at the top of inclined plane=kinetic energy at the bottom of inclined plane=mgh=12*9.8*h

Answer 4

Potential energy of spring=kinetic energy of block

Answer 5

h=?

Answer 6

At last u hv to find out the length of inclined plane

Answer 7

MgHsin30 ⏰=0.5×10^4×4^2

Answer 8

@sikinder is this formula above written right which as been suggested

Answer 9

but the ans is totally diff. the h i get is 0.136m

Answer 10

i was thinking that ur calculation method is right @DominantVampire

Answer 11

is it?

Answer 12

yes i think @DominantVampire

Answer 13

alright. thankyou.