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) ` )

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) )

In fig. both pulleys and the string are massless and all the surfaces are frictionless. Given m_(1)=1kg, m_(2)= 2kg, m_(3)=3kg . The acceleration of m_(1) is

In fig. both pulleys and the string are massless and all the surfaces are frictionless. Given m_(1)=1kg, m_(2)= 2kg, m_(3)=3kg . The acceleration of m_(3) is

In the situation shown in figure F=500 newton applied on the pulley m_(1)=50 kg and m_(2) = 10 kg and pulley and strings are massless and frictionless. Then the acceleration of the pulley is : [g = 10 m//s^(2)]

Consider the situation shown in figure .Both the pulleys and the string and light and all the surfaces are smooth. (a) Find the acceleration of 1kg block. (b) Find the tension in the string. (g = 10 m//s^(2)) .

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

In the arrangment shown pulley is ideal and string is massless. 5kg block is moving on the smooth surface. When theta is 37^(@) , then acceleration of 5kg and 10 kg blocks are a_(1) and a_(2) while tension in string is T then : ( g =10 m//s^(2) ) :

In fig. both pulleys and the string are massless and all the surfaces are frictionless. Given m_(1)=1kg, m_(2)= 2kg, m_(3)=3kg . Find the tension in the string

In the system shown in figure m_(B)=4kg , and m_(A)=2kg . The pulleys are massless and friction is absent everywhere. The acceleration of block A is.