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|dw:1360985504635:dw|
a spring with k  170 N/m is at the top of a frictionless incline of angle 37.0┬░.The lower end of the incline is distance D 1.00 m from the end of the spring, which is at its relaxed length. A 2.00 kg canister is pushed against the spring until the spring is compressed 0.200 m and released from rest. (a) What is the speed of the canister at the instant the spring returns to its relaxed length (which is when the canister loses contact with the spring)? (b) What is the speed of the canister when it reaches the lower end of the incline?
what i did is |dw:1360985802827:dw|

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Other answers:

use energy conservation in both cases. when compressed , E= kx^2/2 + mgh x=0.2m When it is at bottom h = (D+x)sin37
for h the total distance id 1m+.2m (compressed distance )
sin37(1.2)=h
yes you are proceeding correctly
if i solve it i got v=4.20m/s
is this right
for (a)
no this will be the answer to part b i.e. the canister has reached the ground
(b) i got |dw:1360986031295:dw|
@Diwakar how can that be part b
you have calculated h corresponding to the lower end of the incline
it doest not matter right because i got height so the box is top so height works
|dw:1360986099560:dw|
oh sorry the height 1.2sin37 is from bottom to the cannister and not to the top of the incline as i have shown in the figure
|dw:1360986365438:dw|
u mean that
yes the black line shows the heigth 1.2 sin37
shouldnt that be 1sin 37 because the spring not compressed
for part a energy conservation will be |dw:1360986436055:dw|
|dw:1360986652778:dw|
u mean that right
okay for part b u need to use h=1.2sin37 and should we consider the part (a) speed as initial speed
|dw:1360986763295:dw|
you may do that here h = 1*sin37
since the body has already covered 0.2m in part a as it loses contact with the spring
for part b u mean
okay so h=1 sin 37 and i can use above equation right with initial speed from part (a)
yes for part b
okay thx very much

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