Question

A block of mass m is suspended from a spring of force constant k. it is held to keep the spring
in its relaxed length as shown in the figure.
(a) The applied force is decreased gradually so that the block moves downward at negligible speed. How far below the initial position will the block stop?
how can I solve it using work energy theorem

Answer 1

Answer:

As spring force is conservative in nature and it is only force acting on the block. So,

ΔKE+ΔU=o

ΔKE=

2

1

m[v

f

2

−v

i

2

]

vi=vm/s and v

f

=0m/s. So,

ΔKE=

2

1

m[o

2

−v

2

]=

2

−mv

2

ΔUU=

2

1

k[x

f

2

−x

i

2

]

Let x

f

=x and x

i

is o. So,

Δu=

2

k

[x

2

−o

2

]=

2

kx

2

Thus, ΔKE=−δu

⇒

2

−mv

2

=

2

kx

2

⇒x=

k

mv

2

The block will come

k

mv

2

distance from equilibrium position.