A block of mass 500 g slides down on a rough incline plane of incline plane of inclination 53 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 the acceleration of the block is zero, Friction Force = Downward Force $= m g sin θ$
$\Rightarrow f = 0.5 \times 10 \times sin 53^{0} = 5 \times \frac{4}{5} = 4 N$

Work Done by Friction $= F . S = 4 \times 2 = 8 J$

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A block of mass 500 g slides down on a rough incline plane of incline plane of inclination 53 with a uniform speed. Find the work done against the friction as the block slides through 2 m. [g=10m/s2]

Since the acceleration of the block is zero, Friction Force = Downward Force =mgsinθ⇒f=0.5×10×sin530=5×45=4NWork Done by Friction=F.S=4×2=8J

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

Since the acceleration of the block is zero, Friction Force = Downward Force $= m g sin θ$
$\Rightarrow f = 0.5 \times 10 \times sin 53^{0} = 5 \times \frac{4}{5} = 4 N$

Work Done by Friction $= F . S = 4 \times 2 = 8 J$