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Work done by force F to move block of ma...

Work done by force F to move block of mass `2 kg` from A to C very slowly is `(76 lambda) J`. Force F is always acting tangential to path. Equation of path AB is `x^(2)=8y` and BC is straight line which tangent on curve AB to Point B (` mu` between block and path ABC is 0.5). Then value of `'lambda'` is `[g =10m//s^(2)]`

Text Solution

Verified by Experts

Slope of line `BC = (dy)/(dx) =(2x)/(8)=(2 xx4)/(8)=1`
`rArr theta = 45^(@)`

If the mass m is taken from A to C slowly work done by friction will always be equal to the `W_(r) = mu mg x`

Now, by `W_("net") = Delta KE =0`
`W_(F)-mg(10+2)-mu mg(10+4)=0`
`rArr W_(F) = 380 = 76 xx 5`
`rArr lamda = 5`
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