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Standard XII

Physics

Newton's Second Law

A system of t...

Question

A system of two blocks is shown in figure. Friction coefficient between 5 kg and 10 kg block is $μ = 0.6$ and between 10 kg and ground is $μ = 0.4$ . What will be the maximum value of force $F$ , which is applied at the lower block so that 5 kg block does not slip over 10 kg block? $(g = 10 m / s^{2})$ The force applied at the upper block is having fixed magnitude of 80 N (both forces start to act simultaneously).

160 N

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250 N

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210 N

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310 N

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Solution

The correct option is D 310 N

$(F + 80 N) - μ_{2} (5 k g + 10 k g) g = (15 k g) a \frac{F + 80 - 60}{15} = a F - μ_{2} (15) \times 10 - μ_{1} (5 \times 10) = 10 \times a \frac{F - 90}{10} = a S o, w e c a n w r i t e \frac{F + 80 - 60}{15} = \frac{F - 90}{10} F = 310 N$

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Similar questions

Q. The figure shows a system of two blocks of

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Coefficient of friction between $5 k g$ and $10 k g$ block is $0.5$ . If friction between them is $20 N$ , the value of force being applied on $5 k g$ (in $N$ ) is (Assume that the floor is frictionless)

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The correct option is D 310 N (F+ 80 N) − μ2(5 kg + 10 kg)g = (15 kg)a F + 80 − 6015 = a F − μ2(15) × 10 − μ1(5×10) = 10 ×aF − 9010 = aSo, we can write F + 80 − 6015 = F − 9010 F = 310 N

The correct option is D 310 N

$(F + 80 N) - μ_{2} (5 k g + 10 k g) g = (15 k g) a \frac{F + 80 - 60}{15} = a F - μ_{2} (15) \times 10 - μ_{1} (5 \times 10) = 10 \times a \frac{F - 90}{10} = a S o, w e c a n w r i t e \frac{F + 80 - 60}{15} = \frac{F - 90}{10} F = 310 N$