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Standard XII

Physics

Free Body Diagrams

A block of ma...

Question

A block of mass 1 kg is kept on incline plane inside lift as shown in figure. If lift starts accelerating up at 2 m/ $s^{2}$ from rest then work done by frictional force in 1 second is (in J)
( g = 10 m/ $s^{2}$ )

Friction between block and plane is 0.8.

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Solution

FBD of block w.r.t lift

$f_{l} = μ N = μ m (a + g) c o s θ = 0.4 \sqrt{3} (a + g)$ ${θ = 30^{\circ}}$
Since $f_{l} > m (a + g) s i n θ = 0.5 (a + g)$
$f = m (a + g) s i n θ = 6 N$
Work $= f s i n θ \times \frac{1}{2} \times a \times 1^{2} = 3$ J

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A block of mass 1 kg is kept on incline plane inside lift as shown in figure. If lift starts accelerating up at 2 m/s2 from rest then work done by frictional force in 1 second is (in J) ( g = 10 m/s2) Friction between block and plane is 0.8.

FBD of block w.r.t lift fl=μN=μm(a+g)cos θ=0.4√3(a+g) {θ=30∘} Since fl>m(a+g)sin θ=0.5(a+g) f=m(a+g)sin θ=6N Work =f sin θ×12×a×12=3 J

A block of mass 1 kg is kept on incline plane inside lift as shown in figure. If lift starts accelerating up at 2 m/ $s^{2}$ from rest then work done by frictional force in 1 second is (in J)
( g = 10 m/ $s^{2}$ )

Friction between block and plane is 0.8.

FBD of block w.r.t lift

$f_{l} = μ N = μ m (a + g) c o s θ = 0.4 \sqrt{3} (a + g)$ ${θ = 30^{\circ}}$
Since $f_{l} > m (a + g) s i n θ = 0.5 (a + g)$
$f = m (a + g) s i n θ = 6 N$
Work $= f s i n θ \times \frac{1}{2} \times a \times 1^{2} = 3$ J