Question

Problem - 5.37 : A mass of 1kg attached to one end of
a string is first lifted up with acceleration 4.9m/s² and
then lowered with same acceleration. What is the
ratio of tension in string in two cases

help me by explaining step by step explanation​

Answer 1

Explanation:

Given,

a

1

=a

2

=4.9m/s

2

g=9.8m/s

2

m=1kg

Case (1):- From free body diagram,

T

1

−mg=ma

1

T

1

=m(a+g)

T

1

=1(4.9+9.8)

T

1

=17.47N

Case (2):- From free body diagram,

mg−T

2

=ma

2

T

2

=m(g−a)

T

2

=1(9.8−4.9)

T

2

=4.9N

Ratio,

T

2

T

1

=

1

3

T

1

:T

2

=3:1

Answer 2

Given,

a

1

=a

2

=4.9m/s

2

g=9.8m/s

2

m=1kg

Case (1):- From free body diagram,

T

1

−mg=ma

1

T

1

=m(a+g)

T

1

=1(4.9+9.8)

T

1

=17.47N

Case (2):- From free body diagram,

mg−T

2

=ma

2

T

2

=m(g−a)

T

2

=1(9.8−4.9)

T

2

=4.9N

Ratio,

T

2

T

1

=

1

3

T

1

:T

2

=3:1