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What's wrong with my approach? If the surface is smooth then work done should be equal to kinetic energy and so the soln.

What's wrong with my approach?
If the surface is smooth then work done should be equal to kinetic energy and so the soln. 

Question Image
Grade:11

2 Answers

abhay gupta
askIITians Faculty 198 Points
4 months ago
Dear student,
Here’s your solution

you can’t use workdone = f*d here because you’re already equated work done = gain in kinetic energy
both are similar, if you proceed your calculation in gain in kinetic energy then you’ll have the formula f*d
628-2354_Screenshot 2021-03-13 135427.png
so you can’t use the resultant formula as here’s initial velocity is 0

go step by step
i.e. W = ½ m(v2– u2) = ½ mv2
then differentiate it.

Askiitians expert
Ankit
Vibekjyoti Sahoo
145 Points
12 days ago
you can’t use workdone = f*d here because you’re already equated work done = gain in kinetic energy
both are similar, if you proceed your calculation in gain in kinetic energy then you’ll have the formula f*d
628-2354_Screenshot 2021-03-13 135427.png
so you can’t use the resultant formula as here’s initial velocity is 0

go step by step
i.e. W = ½ m(v2– u2) = ½ mv2
then differentiate it.

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