If a block moving up an inclined plane at 30^ | vedclass

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(B) Given: Initial velocity $u = 5 \, m/s$,final velocity $v = 0 \, m/s$,time $t = 0.5 \, s$,and angle of inclination $	heta = 30^{\circ}$.Using the

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$0.6$

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(B) Given: Initial velocity $u = 5 \, m/s$,final velocity $v = 0 \, m/s$,time $t = 0.5 \, s$,and angle of inclination $	heta = 30^{\circ}$.Using the first equation of motion,$v = u + at$,where $a$ is the retardation:$0 = 5 + a(0.5) \implies a = -10 \, m/s^2$.The magnitude of retardation is $10 \, m/s^2$.When a block moves up an inclined plane,the forces acting against the motion are the component of gravity $(mg \sin 	heta)$ and the frictional force $(f = \mu N = \mu mg \cos 	heta)$.The total retardation $a$ is given by: $ma = mg \sin 	heta + \mu mg \cos 	heta$.Dividing by $m$: $a = g(\sin 	heta + \mu \cos 	heta)$.Substituting the values $(g = 10 \, m/s^2)$:$10 = 10(\sin 30^{\circ} + \mu \cos 30^{\circ})$$1 = 0.5 + \mu(\frac{\sqrt{3}}{2})$$0.5 = \mu(0.866)$$\mu = \frac{0.5}{0.866} \approx 0.577$.Rounding to the nearest option,$\mu \approx 0.6$.

If a block moving up an inclined plane at $30^{\circ}$ with a velocity of $5 \, m/s$ stops after $0.5 \, s$,then the coefficient of friction will be nearly:

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