Question

(c) 91
SM) 25
42. The weight of a man in a lif moving upwards with an
acceleration 'a' is 600 N. When the lift moves downwards
with the same acceleration his weight is found to be 360 N.
The real weight of the man is
(a) 380 N
(b) 600 N
(c) 480 N
(d) 700 N
​

Answer 1

Answer:

480 N

Explanation:

600n=m(g+a)

360N=m(g-a)

g=10

on solving both equations(equate m) we get

600-60a=360=36a

a=2.5

substitute in any eq then we get m=48

since weight= m*g = 48*10 = 480N

Answer 2

The real weight of the man is 480 N.

Explanation:

1. Given data

  Gravity acceleration $\mathbf{(g)=10 \ \frac{m}{s^{2}}}$

  Apparent weight of when lift moving in upward direction $\mathbf{(W_{u})=600 \ N}$

  Apparent weight of when lift moving in downward direction $\mathbf{(W_{d})=360 \ N}$

2. Let

   Real mass of man =M  (kg)

   Real weight of man = W = Mg  (N)

   Acceleration of lift in upward and downward direction $\mathbf{=a \ \left ( \frac{m}{s^{2}} \right )}$

3.  Net acceleration on man when lift moving upward = g+a

    So

    Weight of man at time of upward motion of lift $\mathbf{(W_{u})=M(g+a)}$

    Means

    $\mathbf{M(g+a)=W_{u}}$

    $\mathbf{M(g+a)=600}$        ...1)

4.  Net acceleration on man when lift moving downward = g-a

    So

    Weight of man at time of downward motion of lift $\mathbf{(W_{d})=M(g-a)}$

    Means

    $\mathbf{M(g-a)=W_{d}}$

    $\mathbf{M(g-a)=360}$       ...2)

5. Equation 1) ÷ Equation 2)

    $\mathbf{\frac{M(g+a)}{M(g-a)}=\frac{600}{360}}$

    ⇒ $\mathbf{\frac{10+a}{10-a}=\frac{5}{3}}$

    On solving above equation ,we get

    $\mathbf{a=2.5 \ \frac{m}{s^{2}}}$           ...3)

6. From equation 1) and equation 3)

   $\mathbf{M(g+a)=600}$

   $\mathbf{M(10+2.5)=600}$

   $\mathbf{M=\frac{600}{12.5}=48 \ kg}$

   So real weight of man (W) = Mg=48×10=480 N