A block is projected upwards on an inclined p | vedclass

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(A) From the Free Body Diagram $(FBD)$,both the gravitational component $(mg \sin 37^{\circ})$ and the frictional force $(f = \mu N = \mu mg \cos 37^{

Vedclass · Accepted Answer

$1.25$

Vedclass · Answer

(A) From the Free Body Diagram $(FBD)$,both the gravitational component $(mg \sin 37^{\circ})$ and the frictional force $(f = \mu N = \mu mg \cos 37^{\circ})$ act downward along the inclined plane.Therefore,the block experiences a constant deceleration $(a)$ while moving up the plane:$ma = mg \sin 37^{\circ} + \mu mg \cos 37^{\circ}$$a = g(\sin 37^{\circ} + \mu \cos 37^{\circ})$Given $g = 10 \ m/s^2$,$\sin 37^{\circ} \approx 0.6$,$\cos 37^{\circ} \approx 0.8$,and $\mu = 0.5$:$a = 10(0.6 + 0.5 	imes 0.8) = 10(0.6 + 0.4) = 10 \ m/s^2$Using the kinematic equation $v^2 = u^2 - 2aS$,where final velocity $v = 0$ and initial velocity $u = 5 \ m/s$:$0 = (5)^2 - 2(10)S$$20S = 25$$S = 1.25 \ m$

$A$ block is projected upwards on an inclined plane of inclination $37^{\circ}$ along the line of greatest slope with a coefficient of friction $\mu = 0.5$ and an initial velocity of $5 \ m/s$. The block first stops at a distance of .......... $m$ from the starting point.

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