A block of mass 0-8 kg is placed on a rough horizontal floor having coefficient of kinetic friction- -0-2- Find the acceleration in m - s 2 of the block- Take g -10 m - s 2

Weight of the block, W=mg=0.8× 10=8 N FBD of the block, From FBD, N=20sin 53^∘+8=24 f=μ N=0.2 × 24=4.8 N And along the horizontal, a=Fm=20cos53^∘ fm a=12 4.80.8 ...

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Standard XII

Physics

Calculating Friction Using Coeffcients

A block of ma...

Question

A block of mass $0.8 kg$ is placed on a rough horizontal floor having coefficient of kinetic friction, $μ = 0.2$ . Find the acceleration (in ${m/s}^{2}$ ) of the block. Take $g = 10 {m/s}^{2}$

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Solution

Weight of the block, $W = m g = 0.8 \times 10 = 8 N$
FBD of the block,
From FBD,
$N = 20 sin 53^{\circ} + 8 = 24$
$f = μ N = 0.2 \times 24 = 4.8 N$
And along the horizontal,
$a = \frac{F}{m} = \frac{20 cos 53^{\circ} - f}{m}$
$⟹ a = \frac{12 - 4.8}{0.8} = 9 {m/s}^{2}$

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A block of mass 0.8 kg is placed on a rough horizontal floor having coefficient of kinetic friction, μ=0.2. Find the acceleration (in m/s2) of the block. Take g=10 m/s2

Weight of the block, W=mg=0.8×10=8 N FBD of the block, From FBD, N=20sin53∘+8=24 f=μN=0.2×24=4.8 N And along the horizontal, a=Fm=20cos53∘−fm ⟹a=12−4.80.8=9 m/s2

A block of mass $0.8 kg$ is placed on a rough horizontal floor having coefficient of kinetic friction, $μ = 0.2$ . Find the acceleration (in ${m/s}^{2}$ ) of the block. Take $g = 10 {m/s}^{2}$

Weight of the block, $W = m g = 0.8 \times 10 = 8 N$
FBD of the block,
From FBD,
$N = 20 sin 53^{\circ} + 8 = 24$
$f = μ N = 0.2 \times 24 = 4.8 N$
And along the horizontal,
$a = \frac{F}{m} = \frac{20 cos 53^{\circ} - f}{m}$
$⟹ a = \frac{12 - 4.8}{0.8} = 9 {m/s}^{2}$