Physics, asked by coolanyupra, 1 year ago

A 60Kg man is inside a lift which is moving up with an acceleration of 2.45 m/sec2. The apparent percentage change in his weight is

Answers

Answered by nirman95
14

Given:

A 60Kg man is inside a lift which is moving up with an acceleration of 2.45 m/s².

To find:

Percentage change in his Weight.

Calculation:

Weight of the man in stationary lift be w1.

 \therefore \: w1 = mg

  =  > \: w1 = 60 \times 10

  =  > \: w1 = 600 \: N

Weight of man in lift accelerating upwards be w2.

 \therefore \: w2 = m(g + 2.45)

 =  >  \: w2 = 60 \times (10+ 2.45)

 =  >  \: w2 = 60 \times ( 12.45)

 =  >  \: w2 = 747 \: N

Change in weight be ∆w.

∆w = 747 - 600 = 147 N

 \sf{\% \: change =  \dfrac{147}{600}  \times 100\%}

 \sf{ =  > \% \: change =  0.245  \times 100\%}

 \sf{ =  > \% \: change =  24.5 \%}

So , final answer is :

 \boxed{ \sf{ \% \: change =  24.5 \%}}

Answered by pillutlavijaykumar20
4

Explanation:

according to the question answer is given

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