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

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

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

Diwakar
 one year ago
Best ResponseYou've already chosen the best response.1yes you are proceeding correctly

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

Diwakar
 one year ago
Best ResponseYou've already chosen the best response.1no this will be the answer to part b i.e. the canister has reached the ground

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

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

Diwakar
 one year ago
Best ResponseYou've already chosen the best response.1you have calculated h corresponding to the lower end of the incline

ksaimouli
 one year 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

Diwakar
 one year ago
Best ResponseYou've already chosen the best response.1oh 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
 one year ago
Best ResponseYou've already chosen the best response.0dw:1360986365438:dw

Diwakar
 one year ago
Best ResponseYou've already chosen the best response.1yes the black line shows the heigth 1.2 sin37

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

Diwakar
 one year ago
Best ResponseYou've already chosen the best response.1for part a energy conservation will be dw:1360986436055:dw

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

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

Diwakar
 one year ago
Best ResponseYou've already chosen the best response.1you may do that here h = 1*sin37

Diwakar
 one year ago
Best ResponseYou've already chosen the best response.1since the body has already covered 0.2m in part a as it loses contact with the spring

ksaimouli
 one year 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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