## DominantVampire one year ago a block of mass 12 kg is released from rest om a friction-less incline of the angle theta=30 below the block is a spring having spring constant 10^4 N/m. the block stops momentarily when it compresses the spring by 4 cm. how far does the block move down the incline from its rest position to this stopping point

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1. DominantVampire

@sleepyjess @pooja195 @Preetha @e.mccormick please i just need help with this Q.

2. DominantVampire

@sikinder @MrNood

3. DominantVampire

is the answer h=1.70m and x=3.4m can you calculate and tell me, ive solved but i only want to confirm

4. DominantVampire

frist mgh=1/2kx^2 the answer of h which i get it 1.70m then sin30=h/x x= 3.4 this is right

5. DominantVampire

@Miracrown @perl

6. anonymous

@DominantVampire ur calculation and the answer is also right

7. IrishBoy123

Quote: "frist mgh=1/2kx^2 the answer of h which i get it 1.70m" if 12KG drops 1.7m vertically, loss in gravitational $$PE = 12(9.8)(1.7) \simeq 200J$$ if that energy is absorbed in spring k = 10^4 N/m you get extension $$\mathcal{*20*} cm$$ ie $$0.5 \times 10^4 \ \times 0.2^2 = 200J$$

8. Michele_Laino

here we have to keep in mind that the loss of potential energy of the block is the work done by that block against the spring, so that work willtransform itself in potential energy of the spring. So we can write: $\Large mgh = \frac{1}{2}k{\delta ^2}$ |dw:1434278336621:dw| where \delta = 4 cm

9. Michele_Laino

now, since: $\Large h = l\sin \theta = l\sin 30 = \frac{l}{2}$ we get: $\Large mg\frac{l}{2} = \frac{1}{2}k{\delta ^2}$ and solving for l, we get: $\Large l = \frac{{k{\delta ^2}}}{{mg}} = \frac{{{{10}^4} \times 16 \times {{10}^{ - 4}}}}{{12 \times 9.81}} \cong 13.6\;{\text{cm}}$

10. DominantVampire

@IrishBoy123

11. DominantVampire

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