KCET · Physics · Laws of Motion
A man weighing \( 60 \mathrm{Kg} \) is in a lift moving down with an acceleration of
\( 1.8 \mathrm{~ms}^{(-2)} \). The force exerted by the floor on him is
- A \( 588 \mathrm{~N} \)
- B \( 480 \mathrm{~N} \)
- C Zero
- D \( 696 \mathrm{~N} \)
Answer & Solution
Correct Answer
(B) \( 480 \mathrm{~N} \)
Step-by-step Solution
Detailed explanation
Given, weight of man \(=60 \mathrm{~kg}\); acceleration \(=1.8 \mathrm{~ms}^{-2}\)
Force exerted by floor on man \(=m(g-a)\)
where \(g\) is acceleration due to gravity \(=9.8 \mathrm{~ms}^{-2}\)
Therefore, force exerted
\(=60(9.8-1.8)=60 \times 8.0=480 \mathrm{~N}\)
Hence, force exerted by floor on man \(=480 \mathrm{~N}\)
Force exerted by floor on man \(=m(g-a)\)
where \(g\) is acceleration due to gravity \(=9.8 \mathrm{~ms}^{-2}\)
Therefore, force exerted
\(=60(9.8-1.8)=60 \times 8.0=480 \mathrm{~N}\)
Hence, force exerted by floor on man \(=480 \mathrm{~N}\)
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