Following FBD diagram, find the magnitude of kinetic friction on block. Take the acceleration due to gravity to be $g$ = $10 m / s^{2} m = 10 k g μ = 0.5$ . Round off your answer to nearest integer.

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Solution

Normal force acting on the block= $N = m g c o s 30^{\circ} = 86.6 N$
Hence the friction force acting is $f = μ N = 0.5 \times 86.6 N = 43.3 N$

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Following FBD diagram, find the magnitude of kinetic friction on block. Take the acceleration due to gravity to be g = 10m/s2 m=10 kg μ=0.5 . Round off your answer to nearest integer.

Normal force acting on the block=N=mg cos30∘=86.6 NHence the friction force acting is f=μN=0.5×86.6N=43.3 N

Following FBD diagram, find the magnitude of kinetic friction on block. Take the acceleration due to gravity to be $g$ = $10 m / s^{2} m = 10 k g μ = 0.5$ . Round off your answer to nearest integer.

Normal force acting on the block= $N = m g c o s 30^{\circ} = 86.6 N$
Hence the friction force acting is $f = μ N = 0.5 \times 86.6 N = 43.3 N$