A block of mass m has initial velocity u having direction towards x axis the block stops after covering distance s causing similar extension in the spring of constant holding k. If mu is the kinetic friction between the block and the surface on which it was moving , the distance s is given by
Answers
Answer:
got a different answer than the published result.
Did I make a mistake!
Explanation:
Assuming that initially the spring is in its equilibrium position.
Initial energy of block is its kinetic energy
=
1
2
m
v
2
Using Law of Conservation of energy:
When the block stops it initial kinetic energy is converted in to mechanical potential energy of spring which gets stretched by distance
S
and remaining energy is spent doing work against force of friction during its movement.
PE of the spring
=
1
2
K
S
2
Force of friction
=
μ
m
g
Work done against force of friction
=
Force
×
distance
=
μ
m
g
×
S
Equating the initial and final energies we have
1
2
m
v
2
=
1
2
K
S
2
+
μ
m
g
S
⇒
K
S
2
+
2
μ
m
g
S
−
m
v
2
=
0
Solving the quadratic in
S
we get
S
=
−
2
μ
m
g
±
√
(
2
μ
m
g
)
2
−
4
×
K
(
−
m
v
2
)
2
K
S
=
1
K
(
−
μ
m
g
±
√
(
μ
m
g
)
2
+
K
m
v
2
)
Ignoring the
−
v
e
root as the movement is in
+
x
direction we get
S
=
1
K
(
√
(
μ
m
g
)
2
+
K
m
v
2
−
μ
m
g
)