in the adjoining figure, if the block starts of moving from
rest, work done by the frictional force from t = 1 sto i = 28
100N
37°
10 kg
(A) 60 J
(C) 180 J
(B) -120 J
(D) -240 J
Answers
Answer:
this is your answer 180J
Answer:
My answer is Wf = -180J
Explanation:
Given:
A block starts moving from rest under the application of an applied force on a rough horizontal surface.
Coefficient of friction between the block and the surface,
Mass of the block, m = 10 kg
We have to find the work done by the frictional force from t = 1 sec to t = 2 sec.
Note : 10 g is gravitational force since m = 10 kg. Gravitational force is equal to mg]
Balancing the forces on the block in the horizontal direction, we get
100cos37 - f = 10a
where,
f: frictional force acting on the block
a: acceleration of the block
100*4N/5-μN' = 10a
[ cos37= 4/5 ]
where, N' is the normal reaction of the surface on the block.
80N-0.5N'=10a ------> I
Also, balancing the forces in the vertical direction, we get
N'+100sin37=10g
[ g=10m/s² ]
Solve it, you'll get,
N'= 40N -------> II
Substituting the value of N' in eq. (I), we get
a = 6m/s²
This is the acceleration of the block.
Now, we have to find the displacement of the block during t = 1 sec to t = 2 sec.
The block has gained some velocity in the first second which will be its initial velocity between t = 1 sec and t = 2 sec.
This velocity is given by
v = u +at
=> v = 6m/s
Now, for the motion from t = 1 sec and t = 2 sec, displacement of the block is given by
S = ut + 1/2 at²
Substitute the found out values,
S =9m
Work done by the frictional force is given by
Wf = - frictional force * displacement
Wf = -f * s
Wf = -μN' * S
Wf = -20 * 9
Wf = -180J
[negative sign is because, the frictional force and displacement are in the opposite direction and therefore work done is negative]
I hope that my answer is clear and you were satisfied with it.
Thanks for asking your questions.
Questions are to be asked and to be solved.
Once again Thanks!
o(^▽^)o