ksaimouli
  • ksaimouli
solve
Physics
chestercat
  • chestercat
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ksaimouli
  • ksaimouli
|dw:1360985504635:dw|
ksaimouli
  • ksaimouli
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?
ksaimouli
  • ksaimouli
what i did is |dw:1360985802827:dw|

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

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