Two mases m1=5kg and m2=10kg connected by an inextensible string over a frictionless pulley .The coefficient of friction of horizontal surface is 0.15.The minimum weight m that should be put on top of m2 to stop the motion is ?
Answers
Answer:
The minimum weight required is 23.3 kg
Explanation:
According to the problem the two masses are m1= 5 kg and m2= 10 kg they are connected by a string over a pulley.
The coefficient friction is 0.15
Let the tension is T
Therefore the force is m1= m1 x g [ where g is acceleration due to gravity= 10m/s]
= 5 x 10 = 50 N
Therefore,
50 – T = 5 × a.....(1)
T – 0.15 (m + 10) g = (10 + m)a ...(2) [ where m is the minimum weight to put on m2]
As a = 0 for rest
Therefore from 1 and 2 we get
50 = 0.15 (m + 10) 10
5 =3/20(m + 10)
100/3= m + 10
m = 23.3 kg
Answer:
Here is the answer..Hope it Helps :-))
Explanation:
m1 = 5 Kg and m2 = 10 kg
As the system is at rest, T= 50N
T−0.15(m+10)g=(10+m)a
a=0, for rest
50=0.15(m+10)×10
5= 20/3 (m+10)
100/3 =m+10
=> m= 100/3-10
m= 100-30/3
m=70/3
i.e m=23.3 kg