A board of mass 2 kg lies on a table with a load of 4 kg on the board, as shown in the figure. The coefficient of friction between the load and the board is μ 1 = 0.2 and that between the board and

Question

A board of mass 2 kg lies on a table with a load of 4 kg on the board, as shown in the figure. The coefficient of friction between the load and the board is $μ_{1} = 0.2$ and that between the board and the table is $μ_{2} = 0.4$ . The minimum force $F$ that must be applied on the board to pull it out from under the load is: (Take $g = 10$ m/s²)

Options:

Answer:

A

Solution:

$F = μ_{1} M g + μ_{2} (m + M) g + m μ_{1} g = 0.2 \times 40 + 0.4 \times 60 + 2 \times 2 = 8 + 24 + 4 = 36 N .$

Option A.

TestHub · Accepted Answer

F = μ 1 ​ M g + μ 2 ​ ( m + M ) g + m μ 1 ​ g = 0.2 × 40 + 0.4 × 60 + 2 × 2 = 8 + 24 + 4 = 36 N . Option A.

Physics - NLM Question with Solution | TestHub

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