A block of mass $250 g$ slides down an incline of inclination $37^{0}$ with a uniform speed. Find the work done against the friction as the block slides through $1.0 m$ .

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Solution

The friction on the block acts in an upward direction.
Write the force balancing equation for the block on the inclined plane:
$f - m g s i n 37 = 0$
$f = m g s i n 37$

Work done on the block is given as:
$W = f d c o s θ$

Since the angle between force and displacement is zero. So,
$W = m g s i n 37 \times d \times c o s 0$

$= 0.250 \times 10 \times \frac{3}{5} \times 1.0$

$= 1.5 J$

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A block of mass 250 g slides down an incline of inclination 370 with a uniform speed. Find the work done against the friction as the block slides through 1.0 m.

The friction on the block acts in an upward direction.Write the force balancing equation for the block on the inclined plane:f−mgsin37=0f=mgsin37Work done on the block is given as:W=fdcosθSince the angle between force and displacement is zero. So, W=mgsin37×d×cos 0=0.250×10×35×1.0=1.5J

A block of mass $250 g$ slides down an incline of inclination $37^{0}$ with a uniform speed. Find the work done against the friction as the block slides through $1.0 m$ .