A block of mass 2kg is released from A on the track that is one quadrant of a circle of radius 1m. It slides down the track and reaches B with a speed of 4 ms⁻¹ and finally stops at C at a distance of 3m from B. The work done against the force of friction is

The correct answer is 20 J.

To find the total work done against the force of friction as the block travels from point A to point C, we can divide the motion into two distinct segments and apply the work-energy theorem to each.

1. Motion from A to B (along the circular track)
The track is one quadrant of a circle of radius $R=1m$. Therefore, the vertical height of A relative to B is:
$h=1m$

Given data for this segment:
Mass of the block, $m=2\mathrm{kg}$
Initial velocity at A, $v_A=0\mathrm{m/s}$ (released from rest)
Final velocity at B, $v_B=4\mathrm{m/s}$
Acceleration due to gravity, $g=10{\mathrm{m/s}}^2$

Applying the Work-Energy Theorem from A to B:
$W_{\mathrm{gravity}}+W_{\mathrm{friction,\; AB}}=K_B-K_A$

Substitute the formulas for gravitational work and kinetic energy:
$mgh+W_{\mathrm{friction,\; AB}}=\frac{1}{2}m{v}_B^2-0$

Substitute the numerical values:
$\left(2\right)\left(10\right)\left(1\right)+W_{\mathrm{friction,\; AB}}=\frac{1}{2}\left(2\right){\left(4\right)}^2$
$20+W_{\mathrm{friction,\; AB}}=16$
$W_{\mathrm{friction,\; AB}}=16-20=-4J$

Thus, the work done against friction during the descent from A to B is:
$W_{\mathrm{against\; friction,\; AB}}=4J$

2. Motion from B to C (along the horizontal track)
The block travels along the horizontal surface and comes to a stop at C.

Given data for this segment:
Initial velocity at B, $v_B=4\mathrm{m/s}$
Final velocity at C, $v_C=0\mathrm{m/s}$

Applying the Work-Energy Theorem from B to C:
$W_{\mathrm{friction,\; BC}}=K_C-K_B$
$W_{\mathrm{friction,\; BC}}=0-\frac{1}{2}m{v}_B^2$
$W_{\mathrm{friction,\; BC}}=-\frac{1}{2}\left(2\right){\left(4\right)}^2=-16J$

Thus, the work done against friction along the horizontal segment BC is:
$W_{\mathrm{against\; friction,\; BC}}=16J$

3. Total work done against friction
Adding the work done against friction in both segments gives:
$W_{\mathrm{total\; against\; friction}}=W_{\mathrm{against\; friction,\; AB}}+W_{\mathrm{against\; friction,\; BC}}$
$W_{\mathrm{total\; against\; friction}}=4J+16J=20J$