If the coefficient of friction of a plane inclined at 45?, is 0.5, then acceleration of a body sliding freely on it is:

Question

If the coefficient of friction of a plane inclined at 45?, is 0.5, then acceleration of a body sliding freely on it is: (A)  (9.8/2√2)m/s2  (B)  (9.8/√2)m/s2  (C)  9.8m/s2  (D)  4.9m/s2

Tardigrade Admin · Accepted Answer

Acceleration of the body on inclined plane. a=g( sin θ -μ cos θ ) =g( sin 45o-μ cos 45o) =g( (1/√2)-0.5(1/√2) ) =(g/√2)(1-0.5)=(g/2√2) =(9.8/2√2)m/s2

Q. If the coefficient of friction of a plane inclined at 45?, is 0.5, then acceleration of a body sliding freely on it is:

$\frac{9.8}{2 2} m / s^{2}$

$\frac{9.8}{2} m / s^{2}$

$9.8 m / s^{2}$

$4.9 m / s^{2}$

Acceleration of the body on inclined plane. $a = g (sin θ - μ cos θ)$ $= g (sin 45^{o} - μ cos 45^{o})$ $= g (\frac{1}{2} - 0.5 \frac{1}{2})$ $= \frac{g}{2} (1 - 0.5) = \frac{g}{2 2}$ $= \frac{9.8}{2 2} m / s^{2}$

Q. If the coefficient of friction of a plane inclined at 45?, is 0.5, then acceleration of a body sliding freely on it is:

22​9.8​m/s2

2​9.8​m/s2

9.8m/s2

4.9m/s2

Acceleration of the body on inclined plane. a=g(sinθ−μcosθ) =g(sin45o−μcos45o) =g(2​1​−0.52​1​) =2​g​(1−0.5)=22​g​ =22​9.8​m/s2

$\frac{9.8}{2 2} m / s^{2}$

$\frac{9.8}{2} m / s^{2}$

$9.8 m / s^{2}$

$4.9 m / s^{2}$

Acceleration of the body on inclined plane. $a = g (sin θ - μ cos θ)$ $= g (sin 45^{o} - μ cos 45^{o})$ $= g (\frac{1}{2} - 0.5 \frac{1}{2})$ $= \frac{g}{2} (1 - 0.5) = \frac{g}{2 2}$ $= \frac{9.8}{2 2} m / s^{2}$