Physics, asked by shanishprasad786, 5 hours ago

A 0.5-kg block slides along a horizontal frictionless surface at 2 m/s. It is brought to rest by compressing a very long spring of spring constant 800 N/m. The maximum spring compression is

Answers

Answered by kp959049
1

Explanation:

As the track AB is frictionless, the block moves this distance without loss in its initial KE=

2

1

mv

2

=

2

1

×0.5×3

2

=2.25J. In the path BD as friction is present, so work done against friction

k

mgs=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=

2

1

×k×x

2

k

mgx

i.e., 0.11=

2

1

×2×x

2

+0.2×0.5×10x

or x

2

x−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 f

L

s

mg

=0.22×0.5×10=1.1N.Since,F<f

L

. 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

Answered by subha2007293
0

Answer:

As the track AB is frictionless, the block moves this distance without loss in its initial KE=

2

1

mv

2

=

2

1

×0.5×3

2

=2.25J. In the path BD as friction is present, so work done against friction

k

mgs=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=

2

1

×k×x

2

k

mgx

i.e., 0.11=

2

1

×2×x

2

+0.2×0.5×10x

or x

2

x−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 f

L

s

mg

=0.22×0.5×10=1.1N.Since,F<f

L

. 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

hope it help you.

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