Question

A force is applied to an initially stationary block of mass 1.75 kg that sits on a horizontal floor. The 53.5 N force is applied at ? = 32 ? angle. The coefficients of friction between the floor and

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Answer 1

As per the given question

As per the given questionApplying the force equilibrium

As per the given questionApplying the force equilibriumN

As per the given questionApplying the force equilibriumNN=mg+Fcos⁡θN=(1.75×9.8)+(53.5cos⁡32∘)N=62.52N

As per the given questionApplying the force equilibriumNN=mg+Fcos⁡θN=(1.75×9.8)+(53.5cos⁡32∘)N=62.52NNow, the maximum static friction

As per the given questionApplying the force equilibriumNN=mg+Fcos⁡θN=(1.75×9.8)+(53.5cos⁡32∘)N=62.52NNow, the maximum static frictionF

As per the given questionApplying the force equilibriumNN=mg+Fcos⁡θN=(1.75×9.8)+(53.5cos⁡32∘)N=62.52NNow, the maximum static frictionFs

As per the given questionApplying the force equilibriumNN=mg+Fcos⁡θN=(1.75×9.8)+(53.5cos⁡32∘)N=62.52NNow, the maximum static frictionFs=

As per the given questionApplying the force equilibriumNN=mg+Fcos⁡θN=(1.75×9.8)+(53.5cos⁡32∘)N=62.52NNow, the maximum static frictionFs=μ

As per the given questionApplying the force equilibriumNN=mg+Fcos⁡θN=(1.75×9.8)+(53.5cos⁡32∘)N=62.52NNow, the maximum static frictionFs=μs

As per the given questionApplying the force equilibriumNN=mg+Fcos⁡θN=(1.75×9.8)+(53.5cos⁡32∘)N=62.52NNow, the maximum static frictionFs=μsN

As per the given questionApplying the force equilibriumNN=mg+Fcos⁡θN=(1.75×9.8)+(53.5cos⁡32∘)N=62.52NNow, the maximum static frictionFs=μsNF

As per the given questionApplying the force equilibriumNN=mg+Fcos⁡θN=(1.75×9.8)+(53.5cos⁡32∘)N=62.52NNow, the maximum static frictionFs=μsNFs

As per the given questionApplying the force equilibriumNN=mg+Fcos⁡θN=(1.75×9.8)+(53.5cos⁡32∘)N=62.52NNow, the maximum static frictionFs=μsNFs=

As per the given questionApplying the force equilibriumNN=mg+Fcos⁡θN=(1.75×9.8)+(53.5cos⁡32∘)N=62.52NNow, the maximum static frictionFs=μsNFs=0.593

As per the given questionApplying the force equilibriumNN=mg+Fcos⁡θN=(1.75×9.8)+(53.5cos⁡32∘)N=62.52NNow, the maximum static frictionFs=μsNFs=0.593×

As per the given questionApplying the force equilibriumNN=mg+Fcos⁡θN=(1.75×9.8)+(53.5cos⁡32∘)N=62.52NNow, the maximum static frictionFs=μsNFs=0.593×62.52

As per the given questionApplying the force equilibriumNN=mg+Fcos⁡θN=(1.75×9.8)+(53.5cos⁡32∘)N=62.52NNow, the maximum static frictionFs=μsNFs=0.593×62.52F

As per the given questionApplying the force equilibriumNN=mg+Fcos⁡θN=(1.75×9.8)+(53.5cos⁡32∘)N=62.52NNow, the maximum static frictionFs=μsNFs=0.593×62.52Fs

As per the given questionApplying the force equilibriumNN=mg+Fcos⁡θN=(1.75×9.8)+(53.5cos⁡32∘)N=62.52NNow, the maximum static frictionFs=μsNFs=0.593×62.52Fs=

As per the given questionApplying the force equilibriumNN=mg+Fcos⁡θN=(1.75×9.8)+(53.5cos⁡32∘)N=62.52NNow, the maximum static frictionFs=μsNFs=0.593×62.52Fs=28.35

As per the given questionApplying the force equilibriumNN=mg+Fcos⁡θN=(1.75×9.8)+(53.5cos⁡32∘)N=62.52NNow, the maximum static frictionFs=μsNFs=0.593×62.52Fs=28.35N

As per the given questionApplying the force equilibriumNN=mg+Fcos⁡θN=(1.75×9.8)+(53.5cos⁡32∘)N=62.52NNow, the maximum static frictionFs=μsNFs=0.593×62.52Fs=28.35NFh=Fsin⁡θFh=53.5sin⁡32∘Fh=28.35N

As per the given questionApplying the force equilibriumNN=mg+Fcos⁡θN=(1.75×9.8)+(53.5cos⁡32∘)N=62.52NNow, the maximum static frictionFs=μsNFs=0.593×62.52Fs=28.35NFh=Fsin⁡θFh=53.5sin⁡32∘Fh=28.35NHence the static friction is greater than the applied force.

As per the given questionApplying the force equilibriumNN=mg+Fcos⁡θN=(1.75×9.8)+(53.5cos⁡32∘)N=62.52NNow, the maximum static frictionFs=μsNFs=0.593×62.52Fs=28.35NFh=Fsin⁡θFh=53.5sin⁡32∘Fh=28.35NHence the static friction is greater than the applied force.Therefore the block will not move

As per the given questionApplying the force equilibriumNN=mg+Fcos⁡θN=(1.75×9.8)+(53.5cos⁡32∘)N=62.52NNow, the maximum static frictionFs=μsNFs=0.593×62.52Fs=28.35NFh=Fsin⁡θFh=53.5sin⁡32∘Fh=28.35NHence the static friction is greater than the applied force.Therefore the block will not moveThat is acceleration is 0.

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Answer 2

Answer:

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