53. Determine the time in which the smaller block reaches other end of bigger block in the figure 10 N 2 kg u=0.3 u=0.0 8 kg L=3.0 m (b) 8 (a) 4s (c) 2.19 s (d) 2.13 s

Determine the time in which the smaller block reaches other end of bigger block in figure

Determine the time in which the smaller block reaches other end of bigger block in figure

Determine the time in which the smaller block reaches other end of bigger block in figure

Find the time taken by the block to reach the bottom of inclined plane. E= 200 hat(i) N/C , M=1kg , q=5mC,g= 10 m/s^2 mu =0.2

5) A constant force F applied to the lower block of mass 15 kg makes it slide between the upper block of mass 5 kg and the table below, as shown. The coefficients of static (us) and kinetic (uk) friction between the lower block and the table are 0.5 and 0.4 respectively and those between the two blocks are 0.3 and 0.1. The accelerations of the upper and the lower blocks are respectively 5 kg 15 kg (a) 1.96 m/s² and 1.96 m/s? (c) 0.98 m/s and 0.49 m/s b) 1.96 m/s and 3.92 m/s? d) 0.98 m/s and 1.96 m/s? 3

In the figure ( g = 10 m//s^(2) ) .Acceleration of 2kg block is :

Both the blocks as shown in figure are given together a horizontal velocity towards right. The acceleration of the centre of mass of the system of block is (m_(A) = 2m_(B) = 2kg) . (A)zero (B) 5//3 m//s^(2) (C) 7//3 m//s^(2) (D) 2 m//s^(2)

A block of mass 8 kg is sliding on a surface inclined at an angle of 45^(@) with the horizontal. Calculate the acceleration of the block. The coefficient of friction between the block and surface is 0.6. (Take g=10m//s^(2) )

Block A has a mass 3kg and is sliding on a rough horizontal surface with a velocity u_A=2m//s when it makes a direct collision with block B, which has a mass of 2kg and is originally at rest. The collision is perfectly elastic. Determine the velocity of each block just after collision and the distance between the blocks when they stop sliding. The coefficient of kinetic friction between the blocks and the plane is mu_k=0.3 ( Take g=10m//s^2 )