Please help me this question. Physics. Thanks!
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Answered by
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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
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
Divyankasc:
Correct! Thanks!!
Could you please help me with my other questions too??
Answered by
0
Answer:
THE ABOVE ANSWER IS CORRECT . PLZ DO NOT REPORT IT
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