Question

Please help me this question. Physics. Thanks!

Answer 1

It is a question of work power energy.

Do not use friction or laws of motion here,please.

See,
kinetic energy at position 1 is 1/2mv1^2=2.25joule.
Kinetic energy at position 2 is 1/2mV2^2=0 joule.

Work done by frictional force before the block stops is =¶mg(BC+x) where ¶ is coefficient of friction

Work done by spring force before the block stops is 1/2kx^2=x^2

From work energy theorem you can say,
2.14+x+x^2=2.25

Solve and you will get x=0.1m

Total distance=AB+BC+x=4.24m.

Hope that helps.
Please let me know my answer is correct or not.

Thanks.
Tripathy

Answer 2

Answer:

THE ABOVE ANSWER IS CORRECT . PLZ DO NOT REPORT IT