Delete
Share
This Question is Closed
ksaimouli
Best Response
You've already chosen the best response.
0
|dw:1360985504635:dw|
ksaimouli
Best Response
You've already chosen the best response.
0
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
Best Response
You've already chosen the best response.
0
what i did is |dw:1360985802827:dw|
Diwakar
Best Response
You've already chosen the best response.
1
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
Best Response
You've already chosen the best response.
0
for h the total distance id 1m+.2m (compressed distance )
ksaimouli
Best Response
You've already chosen the best response.
0
sin37(1.2)=h
Diwakar
Best Response
You've already chosen the best response.
1
yes you are proceeding correctly
ksaimouli
Best Response
You've already chosen the best response.
0
if i solve it i got v=4.20m/s
ksaimouli
Best Response
You've already chosen the best response.
0
is this right
ksaimouli
Best Response
You've already chosen the best response.
0
for (a)
Diwakar
Best Response
You've already chosen the best response.
1
no this will be the answer to part b i.e. the canister has reached the ground
ksaimouli
Best Response
You've already chosen the best response.
0
(b) i got |dw:1360986031295:dw|
ksaimouli
Best Response
You've already chosen the best response.
0
@Diwakar how can that be part b
Diwakar
Best Response
You've already chosen the best response.
1
you have calculated h corresponding to the lower end of the incline
ksaimouli
Best Response
You've already chosen the best response.
0
it doest not matter right because i got height so the box is top so height works
Diwakar
Best Response
You've already chosen the best response.
1
|dw:1360986099560:dw|
Diwakar
Best Response
You've already chosen the best response.
1
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
Best Response
You've already chosen the best response.
0
|dw:1360986365438:dw|
ksaimouli
Best Response
You've already chosen the best response.
0
u mean that
Diwakar
Best Response
You've already chosen the best response.
1
yes the black line shows the heigth 1.2 sin37
ksaimouli
Best Response
You've already chosen the best response.
0
shouldnt that be 1sin 37 because the spring not compressed
Diwakar
Best Response
You've already chosen the best response.
1
for part a energy conservation will be |dw:1360986436055:dw|
ksaimouli
Best Response
You've already chosen the best response.
0
|dw:1360986652778:dw|
ksaimouli
Best Response
You've already chosen the best response.
0
u mean that right
ksaimouli
Best Response
You've already chosen the best response.
0
okay for part b u need to use h=1.2sin37 and should we consider the part (a) speed as initial speed
ksaimouli
Best Response
You've already chosen the best response.
0
|dw:1360986763295:dw|
Diwakar
Best Response
You've already chosen the best response.
1
you may do that
here h = 1*sin37
Diwakar
Best Response
You've already chosen the best response.
1
since the body has already covered 0.2m in part a as it loses contact with the spring
ksaimouli
Best Response
You've already chosen the best response.
0
for part b u mean
ksaimouli
Best Response
You've already chosen the best response.
0
okay so h=1 sin 37 and i can use above equation right with initial speed from part (a)
ksaimouli
Best Response
You've already chosen the best response.
0
@Diwakar
Diwakar
Best Response
You've already chosen the best response.
1
yes for part b
ksaimouli
Best Response
You've already chosen the best response.
0
okay thx very much