A block of mass 250 g slides down an incline of inclination 37^ ^ ^...

Since the block moves with a uniform speed, its acceleration is zero. Thus, the net force acting on the block along the incline is zero.

The component of gravitational force down the incline is balanced by the frictional force acting up the incline.

$f = mg \sin \theta$

Work done against friction as the block moves a distance $d$ is given by:

$W = f \times d = mg \sin \theta \times d$

Substituting the given values ($m = 0.25 \text{ kg}$, $g = 10 \text{ m/s}^2$, \theta = 37^{}^{{}^{\circ}}, $d = 1.0 \text{ m}$):

W = 0.25 \times 10 \times \sin(37^{}^{{}^{\circ}}) \times 1.0

$W = 2.5 \times \dfrac{3}{5} = 1.5 \text{ J}$