A mass of 2 kg rests on an inclined planes of gradient 1 in 8 as shown- The work done in sliding it from the lowest point of the inclined to the highest point- the coefficient of friction is 0-2- is-19-2 J192 J48 J52 J

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W-mgsin-x3B8-x3BC-mgcos-x3B8-S-sin-x3B8-1-8-cos-x3B8-x221A-63-x221A-64-x2248-1-2-xD7-9-8-xD7-18-0-2-xD7-2-xD7-9-8-xD7-1-xD7-8-52J

A mass of $2 k g$ rests on an inclined planes of gradient $1$ in $8$ as shown. The work done in sliding it from the lowest point of the inclined to the highest point, if the coefficient of friction is $0.2$ , is-

$19.2 J$
$192 J$
$48 J$
$52 J$

$W = (m g sin θ + μ m g cos θ) S$
$(sin θ = 1 / 8, cos θ = \frac{\sqrt{6} 3}{\sqrt{6} 4} \approx 1)$
$= (2 \times 9.8 \times \frac{1}{8} + 0.2 \times 2 \times 9.8 \times 1) \times 8$
$= 52 J$

A mass of 2 kg rests on an inclined planes of gradient 1 in 8 as shown. The work done in sliding it from the lowest point of the inclined to the highest point, if the coefficient of friction is 0.2, is-19.2 J192 J48 J52 J

W=(mgsinθ+μmgcosθ)S(sinθ=1/8,cosθ=√63√64≈1)=(2×9.8×18+0.2×2×9.8×1)×8=52J

A mass of $2 k g$ rests on an inclined planes of gradient $1$ in $8$ as shown. The work done in sliding it from the lowest point of the inclined to the highest point, if the coefficient of friction is $0.2$ , is-

$19.2 J$
$192 J$
$48 J$
$52 J$

$W = (m g sin θ + μ m g cos θ) S$
$(sin θ = 1 / 8, cos θ = \frac{\sqrt{6} 3}{\sqrt{6} 4} \approx 1)$
$= (2 \times 9.8 \times \frac{1}{8} + 0.2 \times 2 \times 9.8 \times 1) \times 8$
$= 52 J$