A block of mass 200 gm is travelling towards right along a rough horizontal surface with a velocity of 2 m/s, while a force of 2N is being applied to it towards the right. Predict the value of frictional force acting on it. Find the work done by the frictional force when the applied force of 2 N ceases to act
Answers
Explanation:
I got:
μ
s
=
0.51
F
=
78.4
N
Explanation:
At the start the
100
N
force is just enough to overcome static friction so we can write:
Force
=
Static Friction
F
=
μ
s
N
where
μ
s
is the coefficient of static friction and
N
=
Normal Reaction that in an horizontal case such this will be equal to the weight of the block, so
N
=
m
g
.
We get:
F
=
μ
s
⋅
m
g
in numbers:
100
=
μ
s
⋅
20
⋅
9.8
μ
s
=
100
20
⋅
9.8
=
0.51
When the movement starts, kinetic friction kicks in and we have that to have uniform motion we need acceleration equal to zero (constant velocity).
We use Newton's Second Law:
Σ
→
F
=
m
→
a
or in our case:
Force
−
Kinetic Friction
=
mass
⋅
acceleration
or
F
−
μ
k
N
=
0
because acceleration has to be zero.
F
−
μ
k
⋅
m
g
=
0
in numbers:
F
−
0.4
⋅
20
⋅
9.8
=
0
F
=
78.4
N
Answer:
Upar wala answer galat hai
Pata nhi kis chutiye ne iss ans. ko likhaa hai