In the system shown in the adjoining figure, the acceleration of the 1 kg mass is `m//s^(2)`. (take `g = 10 m//s^(2)`)

Two blocks of equal masses m are released from the top of a smooth fixed wedge as shown in the figure. Find the magnitude of the acceleration of the centre of mass of the two blocks in m//s^(2) if g =10 m//s^(2)

The fricition coefficient between the table and block shown in fig. is 0.2 find the acceleration of the system (Take g=10 m//s^(2) )

The fricition coefficient between the table and block shown in fig. is 0.2 find the acceleration of the system (Take g=10 m//s^(2) )

The fricition coefficient between the table and block shown in fig. is 0.2 find the acceleration of the system (Take g=10 m//s^(2) )

Find the acceleration of the two blocks shown in figure-2.101.Take g=10m/s2

In the figure shown, the acceleration of block of mass m_(2) is 2m//s^(2) . The acceleration of m_(1) :

In the figure shown, the acceleration of block of mass m_(2) is 2m//s^(2) . The acceleration of m_(1) :

In the figure shown, find acceleration of the system and tenslons T_(1),T_(2) and T_(3) (Take = 10 m//s^(2))

In the figure shown, find acceleration of the system and tenslons T_(1),T_(2) and T_(3) (Take = 10 m//s^(2))

Pulleys are ideal and string are massless. The masses of blocks are m_(1) = 4kg and m_(2) = 1kg as shown.If all surfaces are smooth then the acceleration of m_(2) in m//s^(2) is ( g =10 m//s^(2) )