On a smooth table two blocks of masses 2.5kg and 1.5kg are placed one over the other as shown in figure. If the coefficient of static friction between two blocks is 0.3, the maximum horizontal force to be applied on the lower block so that the two blocks move together is (g =10 ms 2)
Answers
Answer: 8N
Explanation:
Frictional force on 1.5 kg mass:
f=μmBg =0.2×1.5×10 =3 Nf=μmBg =0.2×1.5×10 =3 N
Acceleration:
a=fmB =31.5 =2 m/s2a=fmB =31.5 =2 m/s2
Maximum force:
Fmax=(mA+mB)a =(2.5+1.5)(2) = 8 NFmax=mA+mBa =2.5+1.52 = 8 N
hope this helps everybody
Given:
Mass of 1st block = 2.5 kg
Mass of 2nd block = 1.5 kg
Static friction between the two blocks = 0.3
To find:
The maximum horizontal force to be applied on the lower block so that the two blocks move together.
Solution:
Frictional force on 1.5 kg mass:
f = μmg
= 0.2×1.5×10 =3 N
Acceleration of the upper block:
a= f / m = 3/ 1.5 = 2 m/s²
If the two blocks have to move together then a force should be applied to prevent this acceleration:
Maximum force:
Fmax = (m1 + m2)*a
=(2.5+1.5)*(2) = 8
Therefore the maximum force that can be applied is 8 N.