Question

A lift is moving downwards with an acceleration of 1.8 m is -2 what is Apparent weight released by a man of mass 50 kg

Answer 1

Solution

The apparent weight released by the man is 400 N

Given

• Acceleration (a) = 1.8 m/s²

• Mass of the man, (M) = 50 Kg

To finD

Apparent Weight of the man

Here,

The system is moving downwards with an acceleration 1.8 m/s²

We conclude,

$\tt Mg > Ma$

Now,

$\tt Mg - Ma = R \\ \\ \longrightarrow \tt R = M(g - a) -------(1)$

Here,

• R is the Apparent Weight
• Mg is Weight
• Ma is the force exerted by the man

Substituting the values,we get :

$\tt R = 50(9.8 - 1.8) \\ \\ \longrightarrow \tt R = (50)(8) \\ \\ \large{\longrightarrow \boxed{\boxed{\tt R = 400 N}}}$

Answer 2

Answer:

400 N.

Explanation:since a & g are given,

                                       mg>ma

                                       mg-ma = R

                                       R=M(g-a)

                                       M= 50 kg

                                       g= 9.8 ms^-2

                                       R = 50 (9.8 - 1.8)

                                       R= 50(8)

                                       R=400N.