Question

A person of mass M kg is standing on the left the lift is moving vertically upward according to VT graph then find out the weight of man at the following instant

Answer 1

Answer:

The weight of the person are 20M, 10M and 9.2M at 1s, 8s and 12 sec respectively

Explanation:

Mass of a person = M kg

If lift has upward acceleration then a person weight will be

$W=m(g-(-a))$

$W=m(g+a)$

If lift has downward acceleration then a person weight will be

$W=m(g-a)$

According to figure,

The velocity is 10 m/s at t = 1 sec .

(I) The person weight will be mg.

t = 1 sec

Using equation of motion

$v = u+at$

$10=0+a$

tex]a = 10\ m/s^2[/tex]

The weight of a person is

$W=m(g+a)$

$W=M(10+10)=20M$

(II). A person is moving with constant velocity at t = 8 sec then the acceleration will be zero,

The weight of the person is

$W = m(g+a)$

Here, a = 0

$W = m(g)$

$W=10M$

(III) According to figure,

The velocity is 10 m/s at t = 1 sec .

The person weight will be mg.

At t = 12 sec

Using equation of motion

$v = u+at$

$10=0+a\times12$

tex]a = -\dfrac{10}{12}\ m/s^2[/tex]

The weight of a person is

$W=m(g+a)$

$W=M(10-0.83)=9.2M$

Hence, The weight of the person are 20M, 10M and 9.2M at 1s, 8s and 12 sec respectively