A body is moving along a rough horizontal surface with an initial velocity 6m/s. If the body comes to rest after travelling 9m, then the coefficient of sliding firction will be
A
0.4
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
0.2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
0.6
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
0.8
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is B0.2 We know that frictional force is given by F=μsN=μs×mg
where μs is the coefficient of sliding friction.
If a is the deceleration of the body, ma=μs×mg⇒μs=ag
Now, by third equation of motion we know that
v2−u2=2as v=0 (as body comes to rest) ⇒a=−u22s=−622×9=−2m/s2 (−ve sign shows that the body is retarding or slowing down)