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Answers
Explanation:
As the track AB is frictionless, the block moves this distance without loss in its initial KE=21mv2=21×0.5×32=2.25J. In the path BD as friction is present, so work done against friction
=μkmgs=0.2×0.5×10×2.14=2.14J
So, at D the KE of the block is=2.25-2.14=0.11J
Now, if the spring is compressed by x
0.11=21×k×x2+μkmgx
i.e., 0.11=21×2×x2+0.2×0.5×10x
or x2x−0.11=0
which on solving gives positive value of x=0.1 m
After moving the distance x=0.1 m the block comes to rest. Now the compressed spring exerts a force:
F=kx=2×0.1=0.2N
On the block while limiting frictional force between block and track is fL=μsmg
=0.22×0.5×10=1.1N.Since,F<fL. The block will not move back. So, the total distance moved by the block
=AB+BD+0.1
=2+2.14+0.1
=4.24m
Answer:
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Explanation:
by the way l m Dipti of 8std age 13...