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An object of mass 2 kg moving with a constant velocity of 10 m s⁻¹ encounters a rough patch where the force of friction on the object is 7 N. At the same time, an additional constant force of 3 N opposing the motion is applied on the object. After entering the rough patch, how much distance does the object travel before coming to rest?
View Answer
Total opposing force = 7 + 3 = 10 N. Acceleration a = F m = -10 2 = -5 m s -2 .
Using v 2 = u 2 + 2as with v = 0, u = 10: 0 = 100 + 2(-5)s , so s = 100 10 = 10 m .
Distance travelled in the rough patch
Add the opposing forces
friction 7 N + extra force 3 N =
10 N
, both against the motion
↓
Find the acceleration
a =
-10
2
= -5
m s
-2
↓
Use
v
2
= u
2
+ 2as
0 = 10
2
+ 2(-5)s
↓
s = 10 m
— the object travels 10 m before coming to rest
