A block of mass $500 g$ slides down a a rough incline plane of inclination $53^{o}$ with a uniform speed. Find the work done against the friction as the block slides through $2 m$ $[g = 10 m / s^{- 2}]$

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Solution

Since it is moving with constant speed work done $= △ K t = 0$
$W_{g r a v i t y} - = - W_{f r i c t i o n}$
$W_{f r i c t i o n} = - W_{g r a v i t y}$
$= M g (H_{2} - H_{1})$
$sin 53 = \frac{H_{2} - H_{1}}{2}$
$H_{2} - H_{1} = \frac{8}{5}$
$\frac{4}{5} = \frac{H_{2} - H_{1}}{2}$
$= \frac{500}{1000} \times 10 \times \frac{8}{5}$
$= 8$ joules

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A block of mass 500g slides down a a rough incline plane of inclination 53o with a uniform speed. Find the work done against the friction as the block slides through 2m [g=10m/s−2]

Since it is moving with constant speed work done =△Kt=0Wgravity−=−WfrictionWfriction=−Wgravity=Mg(H2−H1)sin53=H2−H12H2−H1=8545=H2−H12=5001000×10×85=8 joules

A block of mass $500 g$ slides down a a rough incline plane of inclination $53^{o}$ with a uniform speed. Find the work done against the friction as the block slides through $2 m$ $[g = 10 m / s^{- 2}]$