A pendulum bob has a speed 3-m-s while passing through its lowest position- length of the pendulum is 0-5-m then its speed when it makes an angle of 60-circ- with the vertical is-2-m-s-1-m-s-4-m-s-3-m-s

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Toppr Admin · Accepted Answer

Since energy is conserved-mu22-mv22-mgh-x21D2-u2-v2-2gh-x21D2-3-2-v2-2-9-8-0-5-x2212-0-5cos60-x21D2-v-2m-s

A pendulum bob has a speed $3 m / s$ while passing through its lowest position, length of the pendulum is $0.5 m$ then its speed when it makes an angle of $60^{\circ}$ with the vertical is
$2 m / s$
$1 m / s$
$4 m / s$
$3 m / s$

Since energy is conserved:
$\frac{m u^{2}}{2} = \frac{m v^{2}}{2} + m g h \Rightarrow u^{2} = v^{2} + 2 g h \Rightarrow {(3)}^{2} = v^{2} + 2 (9.8) (0.5 - 0.5 cos 60) \Rightarrow v = 2 m / s$

A pendulum bob has a speed 3m/s while passing through its lowest position, length of the pendulum is 0.5m then its speed when it makes an angle of 60∘ with the vertical is2m/s1m/s4m/s3m/s

Since energy is conserved:mu22=mv22+mgh⇒u2=v2+2gh⇒(3)2=v2+2(9.8)(0.5−0.5cos60)⇒v=2m/s

A pendulum bob has a speed $3 m / s$ while passing through its lowest position, length of the pendulum is $0.5 m$ then its speed when it makes an angle of $60^{\circ}$ with the vertical is
$2 m / s$
$1 m / s$
$4 m / s$
$3 m / s$

Since energy is conserved:
$\frac{m u^{2}}{2} = \frac{m v^{2}}{2} + m g h \Rightarrow u^{2} = v^{2} + 2 g h \Rightarrow {(3)}^{2} = v^{2} + 2 (9.8) (0.5 - 0.5 cos 60) \Rightarrow v = 2 m / s$