A block of mass 10 kg, initially at rest, ma | vedclass

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Question

(A) The acceleration $a$ of a block sliding down an inclined plane is given by the formula:$a = g \sin 	heta - \mu g \cos 	heta$Given: mass $m = 10

Vedclass · Accepted Answer

$7 \sqrt{2} \ m$

Vedclass · Answer

(A) The acceleration $a$ of a block sliding down an inclined plane is given by the formula:$a = g \sin 	heta - \mu g \cos 	heta$Given: mass $m = 10 \ kg$, angle $	heta = 45^{\circ}$, coefficient of kinetic friction $\mu = 0.3$, time $t = 2 \ s$, and acceleration due to gravity $g = 10 \ m/s^2$.Substituting these values into the acceleration formula:$a = 10 \sin 45^{\circ} - 0.3 	imes 10 \cos 45^{\circ}$$a = 10 	imes \frac{1}{\sqrt{2}} - 3 	imes \frac{1}{\sqrt{2}} = \frac{7}{\sqrt{2}} \ m/s^2$Using the second equation of motion for distance $s$ with initial velocity $u = 0$:$s = ut + \frac{1}{2}at^2$$s = 0 	imes 2 + \frac{1}{2} 	imes \left( \frac{7}{\sqrt{2}} ight) 	imes (2)^2$$s = \frac{1}{2} 	imes \frac{7}{\sqrt{2}} 	imes 4 = \frac{14}{\sqrt{2}} = 7 \sqrt{2} \ m$Thus, the distance travelled is $7 \sqrt{2} \ m$.

$A$ block of mass $10 \ kg$, initially at rest, makes a downward motion on a $45^{\circ}$ inclined plane. The distance travelled by the block after $2 \ s$ is (Assume the coefficient of kinetic friction to be $0.3$ and $g=10 \ m/s^2$)

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