
In frame of block of mass M moving with acceleration a.

m1a – T = 2m1
10a – T = 20 …(i)

T – m2g = 2m2
T – 200 = 40
T = 240 …(ii)
From equation 1 and 2
10a = 260
or, a = 26 m/s2
for block
F = (M + m2)a
F= 120 × 26
F= 3120 N
So, the correct option is (C): 3120 N

Potential energy (V) versus distance (x) is given by the graph. Rank various regions as per the magnitudes of the force (F) acting on a particle from high to low. 
In the given figure, the blocks $A$, $B$ and $C$ weigh $4\,\text{kg}$, $6\,\text{kg}$ and $8\,\text{kg}$ respectively. The coefficient of sliding friction between any two surfaces is $0.5$. The force $\vec{F}$ required to slide the block $C$ with constant speed is ___ N.
(Given: $g = 10\,\text{m s}^{-2}$) 
Method used for separation of mixture of products (B and C) obtained in the following reaction is: 
Read More: Work and Energy