A block of mass 2 kg slides down an inclined plane of inclination 30- The coefficient of friction between the block and plane is 0-5- The contact force between block and plank is-

The correct option is D 5 √(15) Nt ...

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Centripetal Acceleration

A block of ma...

Question

A block of mass $2 k g$ slides down an inclined plane of inclination $30^{\circ}$ . The coefficient of friction between the block and plane is $0.5$ . The contact force between block and plank is:

20 N t

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10 \sqrt{3} N t

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5 \sqrt{7} N t

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5 \sqrt{15} N t

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Solution

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Similar questions

2 kg

30^{\circ}

0.5

30^{0}

2 kg

30^{\circ}

0.5

m

θ

μ

30^{\circ}

4 kg

μ_{1} = 0.3

2 kg

μ_{2} = 0.20

2 kg

g = 10 {m/s}^{2}

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