Question

Three blocks A, B and C are kept as shown in the figure.The coefficient of friction between A and B is 0.2, B and C is 0.1, C and ground is 0.0. The mass of A, B and C are 3 kg, 2 kg and 1 kg respectively. A is given a horizontal velocity 10 m/s. A, B and C always remain in contact i.e. lies as in figure. The total work done by the friction will be:

Answer 1

Answer:

lies as in figure . the total work done by the friction will be . 1kg respectively . a,b and c alwayas remain in contact i.e lies as in figure . the total work done by the friction will be maiter.

Answer 2

Total work done by the friction will be 0.5 J

Given:

Mass of A = 3 kg

Mass of B = 2 kg

Mass of C = 1 kg

Horizontal velocity of A = 10 m/s

To find:

Total work done by the friction = ?

Formula used:

Momentum is given by the formula:

$\bold{P=mv}$

$\bold{K.E=\frac{1}{2}mv^2}$

Where,

m = Mass

v = Velocity

Solution:

Initial momentum of the system is:

$\bold{P_1=3\times10=30 \ kg.m/s}$

Since, there is no external energy acting on the system. The initial momentum of the system remains conserved.

So, the velocity after some time is:

$\bold{P_2=30=6\times v}$

Thus, the final velocity is:

$\bold{\therefore v= 5 \ m/s}$

Now, the block as moving as a system. So, the mass is taken as 6 kg.

According to conservation of energy,

Initial kinetic energy + Frictional work done = Final kinetic energy

$\bold{(\frac{1}{2}\times3\times10^2)+FWD=(\frac{1}{2}\times6\times5^2)}$

$\bold{FWD=\frac{2\times25}{100}}$

$\bold{\therefore FWD=0.5 \ J}$