Answered

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1. A 1. 2 kg stone is tied to a string and swung in a vertical circle with a radius of 0. 75 m. The

string can withstand a tension of 40. 0 N.

At what maximum speed can the stone move at the bottom of its path without the string

breaking?

0. 95 m/s

4. 2 m/s

5. 0 m/s

5. 7 m/s.

Sagot :

MrChrisidn

[tex] \underline{\underline{\large{\orange{\cal{ ✒ GIVEN:}}}}} [/tex]

[tex]\bullet \: \: \rm{Mass \: of \: the \: stone, m = \: 1.2 \: kg}[/tex]

[tex]\bullet \: \: \rm{Radius, r = 0.75 m}[/tex]

[tex]\bullet \: \: \small{\rm{Tension \: in \: the \: string, T = 40 \: N}}[/tex]

[tex]\bullet \: \: \rm{Acceleration \: due \: to \: gravity=9.8 m/s^{2}}[/tex]

[tex] \underline{\underline{\large{\orange{\cal{REQUIRED:}}}}} [/tex]

At what maximum speed can the stone move at the bottom of its path without the string breaking?

[tex] \underline{\underline{\large{\orange{\cal{SOLUTION:}}}}} [/tex]

At the bottom of the path, the tension in the string must provide the centripetal force in addition to supporting the weight of the stone.

Therefore, the tension T is the sum of the centripetal force [tex]\rm{F_c}[/tex] and the gravitational force [tex]\rm{F_g}[/tex]

[tex]\large{\bullet{\cal{FORMULA'S:}}}[/tex]

[tex]\boxed{\large{\bm{\red{T=F_{c} + F_{g}}}}}[/tex]

[tex]\boxed{\large{\bm{\red{T= \dfrac{ {mv}^{2} }{r} + mg}}}}[/tex]

Now let's [tex]\tt{\purple{substitute}}[/tex] the parameters in the given formula:

[tex]\tt{40 = \dfrac{ {1.2v}^{2} }{0.75} + 1.2(9.8)}[/tex]

[tex]\tt{ \dfrac{ {1.2v}^{2} }{0.75} = 40 - 11.76}[/tex]

[tex]\tt{1.2 {v}^{2} = 28.24 \times 0.75}[/tex]

[tex]\tt{ {v}^{2} = \dfrac{21.28}{1.2} }[/tex]

[tex]\tt{ {v}^{2} = 17.65}[/tex]

[tex]\tt{v = \sqrt{17.65} }[/tex]

[tex]\large{\tt{\purple{40.20 m/s}}}[/tex]

Final Answer:

The stone can move at the bottom of its path without the string breaking at a maximum speed of 4.2 m/s

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