A man travels along a straight horizontal road carrying a block of...

Mass of the block, $m = 2 \text{ kg}$

Acceleration of the block, $a = 0.5 \text{ m/s}^2$

Displacement of the block, $s = 40 \text{ m}$

The force exerted by the man on the block has two components: a vertical component to balance the weight of the block and a horizontal component to provide the acceleration.

The vertical force is $F_y = mg$. The work done by this force is zero because it is perpendicular to the horizontal displacement.

The horizontal force is $F_x = ma$. The work done by this force is given by $W = F_x s = mas$.

Substituting the given values:

$W = 2 \times 0.5 \times 40 = 40 \text{ J}$

Alternatively, using the work-energy theorem, the work done by the net force is equal to the change in kinetic energy of the block:

$W = \Delta K = \dfrac{1}{2}mv^2 - \dfrac{1}{2}mu^2$

Using the equation of motion $v^2 - u^2 = 2as$:

$W = \dfrac{1}{2}m(2as) = mas = 2 \times 0.5 \times 40 = 40 \text{ J}$