The door of the refrigerator has an area of 0.72 m 2 . Show that the minimum forceneeded to open the refrigerator door is about 4 kN.
Question
The door of the refrigerator has an area of 0.72 m 2 . Show that the minimum forceneeded to open the refrigerator door is about 4 kN.
Solution
To solve this problem, we need to use the formula for pressure which is P = F/A, where P is the pressure, F is the force, and A is the area.
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We know that the pressure inside a refrigerator is slightly higher than the atmospheric pressure. The atmospheric pressure is about 101325 Pa (Pascal). Let's assume that the pressure inside the refrigerator is about 101325 Pa + 100 Pa = 101425 Pa.
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We know that the area of the refrigerator door is 0.72 m^2.
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We can rearrange the formula for pressure to solve for force: F = P * A.
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Substituting the given values into the formula gives us F = 101425 Pa * 0.72 m^2 = 73026 N.
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However, the question asks for the answer in kilonewtons (kN), not newtons (N). To convert from N to kN, we divide by 1000: 73026 N / 1000 = 73.026 kN.
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Therefore, the minimum force needed to open the refrigerator door is about 73 kN, not 4 kN.
Please note that the answer is significantly higher than 4 kN. The discrepancy might be due to different assumptions about the pressure inside the refrigerator or a mistake in the problem statement.
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