## ksaimouli 2 years ago solve

1. ksaimouli

|dw:1360985504635:dw|

2. 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?

3. ksaimouli

what i did is |dw:1360985802827:dw|

4. Diwakar

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

5. ksaimouli

for h the total distance id 1m+.2m (compressed distance )

6. ksaimouli

sin37(1.2)=h

7. Diwakar

yes you are proceeding correctly

8. ksaimouli

if i solve it i got v=4.20m/s

9. ksaimouli

is this right

10. ksaimouli

for (a)

11. Diwakar

no this will be the answer to part b i.e. the canister has reached the ground

12. ksaimouli

(b) i got |dw:1360986031295:dw|

13. ksaimouli

@Diwakar how can that be part b

14. Diwakar

you have calculated h corresponding to the lower end of the incline

15. ksaimouli

it doest not matter right because i got height so the box is top so height works

16. Diwakar

|dw:1360986099560:dw|

17. Diwakar

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

18. ksaimouli

|dw:1360986365438:dw|

19. ksaimouli

u mean that

20. Diwakar

yes the black line shows the heigth 1.2 sin37

21. ksaimouli

shouldnt that be 1sin 37 because the spring not compressed

22. Diwakar

for part a energy conservation will be |dw:1360986436055:dw|

23. ksaimouli

|dw:1360986652778:dw|

24. ksaimouli

u mean that right

25. ksaimouli

okay for part b u need to use h=1.2sin37 and should we consider the part (a) speed as initial speed

26. ksaimouli

|dw:1360986763295:dw|

27. Diwakar

you may do that here h = 1*sin37

28. Diwakar

since the body has already covered 0.2m in part a as it loses contact with the spring

29. ksaimouli

for part b u mean

30. ksaimouli

okay so h=1 sin 37 and i can use above equation right with initial speed from part (a)

31. ksaimouli

@Diwakar

32. Diwakar

yes for part b

33. ksaimouli

okay thx very much