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Question

# A block of mass m slides along a floor while a force of magnitude F is applied to it at an angle θ as shown in figure. The coefficient of kinetic friction is μK. Then, the block's acceleration ′a′ is given by: (g is acceleration due to gravity)

A
FmcosθμK(gFmsinθ)
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B
FmcosθμK(g+Fmsinθ)
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C
Fmcosθ+μK(gFmsinθ)
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D
FmcosθμK(gFmsinθ)
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Solution

## The correct option is A Fmcosθ−μK(g−Fmsinθ)Drawing the FBD of the block. Along vertical direction: N=mg−Fsinθ ...(1) & along horizontal direction, Fcosθ−μKN=ma ...(2) Substituting the value of N from eq. (1) in eq. (2), ⇒Fcosθ−μK(mg−Fsinθ)=ma ⇒a=Fmcosθ−μK(g−Fmsinθ) Hence, option (a) is correct.

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