दिया गया `k_1=1500Nm^-1`, `k_2=500Nm^-1`, `m_1=2kg`, `m_2=1kg`.

Given k_1=1500Nm^-1 , k_2=500Nm^-1 , m_1=2kg , m_2=1kg . Find: a. potential energy stored in the spring in equilibrium, and b. work done in slowly pulling down m_2 by 8cm .

m_(A) = 1 kg, m_(B) = 2 kg, m_(C) = 10 kg

1m=___nm

Two blocks of masses m_1 and m_2 , which are connected with a light string, are placed over a friction less pulley. This set up is placed over a weighing machine, as shown. Three combination of masses m_1 and m_2 are used, in first case m_1=6kg and m_2=2kg , in second case m_1=5kg and m_2=3kg and in third case m_1=4kg and m_2=4kg. masses are held stationary initially and then released if the readings of the weighing machine after the release in three cases are W_1 , W_2 and W_3 respectively then:

In the figure shown, the string and pulley are massless. The inclined plane is fixed and has coefficients of friction mu_(s)=0.25 and mu_(k)=0.2 . Consider two cases: Case I : m_(1)=2kg m_(2)=1.5 kg theta=37^(@) , "Case II":m_(1)=3kg m_(2)=2 kg theta =53^(@) Take g=10 m//s^(2), sin 37^(@)=3/5, sin 53^(@)=4/5

Find M (in kg ) for which m_(2) will start sliding over m_(1) .Table on which m_(1) and m_(2) are placed is smooth, mu is the friction coefficient between m_(1) and m_(2) . (Given : m_(1) =2kg, m_(2) =1kg, mu =0.5, g=10 m/s^(2) )

In an elevator a system is arranged as shown in figure. Initially elevator is at rest and the system is in equilibrium with middle spring unstretched. When the elevator accelerated upwards, it was found that for the new equilibrium position (with respect to lift), the further extension i the top spring is 1.5 times that of the further compression in the bottom spring, irrespective of the value of acceleration (a) Find the value of (m_1)/(m_2) in terms of spring constants for this happen. (b) if k_1=k_2=k_3=500(N)/(m) and m_1=2 kg and acceleration of the elevator is 2.5(m)/(s^2) , find the tension in the middle spring in the final equilibrium with respect to lift.

Two consecutive irreversible fierst order reactions can be represented by Aoverset(k_(1))(rarr)Boverset(k_(2))(rarr)C The rate equation for A is readily interated to obtain [A]_(t)=[A]_(0).e^(-k_(1^(t))) , and [B]_(t)=(k_(1)[A]_(0))/(k_(2)-k_(1))[e^(-k_(1)^(t))-e^(-k_(2)^(t))] When k_(1)=1s^(-1) and k_(2)=500s^(-1) , select most appropriate graph

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