A body of mass m kg slides from rest along t | vedclass

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(B) Apply the Work-Energy Theorem $(WET)$ from point $A$ to point $B$: $W_{mg} = K_{B} - K_{A}$ Since the path is frictionless,the work done by gravit

Vedclass · Accepted Answer

$21.9$

Vedclass · Answer

(B) Apply the Work-Energy Theorem $(WET)$ from point $A$ to point $B$: $W_{mg} = K_{B} - K_{A}$ Since the path is frictionless,the work done by gravity is equal to the change in potential energy: $mg 	imes h = \frac{1}{2} mv_{B}^2 - 0$ From the geometry,the vertical height $h$ dropped by the body from $A$ to $B$ is $h = R \sin(45^{\circ}) + R = \frac{R}{\sqrt{2}} + R$. Substituting the values: $mg 	imes R \left( \frac{1}{\sqrt{2}} + 1 ight) = \frac{1}{2} mv_{B}^2$ $gR \left( \frac{1 + \sqrt{2}}{\sqrt{2}} ight) = \frac{1}{2} v_{B}^2$ $v_{B}^2 = 2gR \left( \frac{1 + 1.4}{1.4} ight) = 2 	imes 10 	imes 14 	imes \left( \frac{2.4}{1.4} ight)$ $v_{B}^2 = 20 	imes 10 	imes 2.4 = 480$ $v_{B} = \sqrt{480} \approx 21.9 \ m/s$. Thus,option $B$ is correct.

$A$ body of mass $m \ kg$ slides from rest along the curve of a vertical circle from point $A$ to $B$ on a frictionless path. The velocity of the body at $B$ is: (Given: $R = 14 \ m$,$g = 10 \ m/s^2$,and $\sqrt{2} = 1.4$) (in $m/s$)

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