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ksaimouli
 3 years ago
solve
ksaimouli
 3 years ago
solve

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ksaimouli
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1360985504635:dw

ksaimouli
 3 years ago
Best ResponseYou've already chosen the best response.0a 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
 3 years ago
Best ResponseYou've already chosen the best response.0what i did is dw:1360985802827:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0use 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
 3 years ago
Best ResponseYou've already chosen the best response.0for h the total distance id 1m+.2m (compressed distance )

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes you are proceeding correctly

ksaimouli
 3 years ago
Best ResponseYou've already chosen the best response.0if i solve it i got v=4.20m/s

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0no this will be the answer to part b i.e. the canister has reached the ground

ksaimouli
 3 years ago
Best ResponseYou've already chosen the best response.0(b) i got dw:1360986031295:dw

ksaimouli
 3 years ago
Best ResponseYou've already chosen the best response.0@Diwakar how can that be part b

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you have calculated h corresponding to the lower end of the incline

ksaimouli
 3 years ago
Best ResponseYou've already chosen the best response.0it doest not matter right because i got height so the box is top so height works

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1360986099560:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh 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
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1360986365438:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes the black line shows the heigth 1.2 sin37

ksaimouli
 3 years ago
Best ResponseYou've already chosen the best response.0shouldnt that be 1sin 37 because the spring not compressed

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0for part a energy conservation will be dw:1360986436055:dw

ksaimouli
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1360986652778:dw

ksaimouli
 3 years ago
Best ResponseYou've already chosen the best response.0okay for part b u need to use h=1.2sin37 and should we consider the part (a) speed as initial speed

ksaimouli
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1360986763295:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you may do that here h = 1*sin37

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0since the body has already covered 0.2m in part a as it loses contact with the spring

ksaimouli
 3 years ago
Best ResponseYou've already chosen the best response.0okay so h=1 sin 37 and i can use above equation right with initial speed from part (a)
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