Question

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

Answer 1

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

Answer 2

@sikinder @MrNood

Answer 3

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

Answer 4

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

Answer 5

@Miracrown @perl

Answer 6

@DominantVampire ur calculation and the answer is also right

Answer 7

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 \(0.5 \times 10^4 \ \times 0.2^2 = 200J\)

Answer 8

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:
\[\Large mgh = \frac{1}{2}k{\delta ^2}\]

where \delta = 4 cm

Answer 9

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:
\[\Large l = \frac{{k{\delta ^2}}}{{mg}} = \frac{{{{10}^4} \times 16 \times {{10}^{ - 4}}}}{{12 \times 9.81}} \cong 13.6\;{\text{cm}}\]

Answer 10

@IrishBoy123

Answer 11

3rd last