Physics, 12.10.2019 00:10 laerica

Ayoung kid of mass m = 36 kg is swinging on a swing. the length from the top of the swing set to the seat is l = 3.5 m. the boy is attempting to swing all the way around in a full circle. what is the minimum speed, in meters per second, the boy must be moving with at the top of the path in order to make a full circle? v = assuming the boy is traveling at the speed found in part (a), what is their apparent weight. wa in newtons, at the top of their path? (at the top. the boy is upside-down.) what is the boy's apparent weight, in newtons, at the bottom of their path if they have the velocity from part (a) at the top?

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Answer from: jalst6084

We have that for the question it can be said that their apparent weight. Wa in newtons, at the top of their path is

W=609.9428N

From the question we are told

A young kid of mass m = 36 kg is swinging on a swing.

The length from the top of the swing set to the seat is L = 3.5 m.

The boy is attempting to swing all the way around in a full circle.

What is the minimum speed, in meters per second, the boy must be moving with at the top of the path in order to make a full circle? v = Assuming the boy is traveling at the speed found in part (a), what is their apparent weight. Wa in newtons, at the top of their path? (At the top. the boy is upside-down.) What is the boy's apparent weight, in newtons, at the bottom of their path if they have the velocity from part (a) at the top?

Assuming value for v=5

Generally the equation for the Weight is mathematically given as

$W=ma_2\\\\W=\frac{mv^2}{L}+mg\\\\W=\frac{36 *5^2}{3.5}+36*9.8$

W=609.9428N

Therefore

Their apparent weight. Wa in newtons, at the top of their path is

W=609.9428N

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Answer from: terryhgivens4294

Explanation:

Given

mass of boy=36 kg

length of swing=3.5 m

Let T be the tension in the swing

At top point $mg-T=\frac{mv^2}{r}$

where v=velocity needed to complete circular path

Th-resold velocity is given by $mg-0=\frac{mv^2}{r}$

$v=\sqrt{gr}=\sqrt{9.8\times 3.5}=5.85 m/s$

So apparent weight of boy will be zero at top when it travels with a velocity of $v=\sqrt{gr}$

To get the velocity at bottom conserve energy at Top and bottom

At top $E_T=mg\times 2L+\frac{mv^2}{2}$

Energy at Bottom $E_b=\frac{mv_0^2}{2}$

Comparing two as energy is conserved

v_0^2=4gl+gl

v_0^2=5gL

$v_0=\sqrt{5gL}=13.09 m/s$

Apparent weight at bottom is given by

$W=\frac{mv_0^2}{L}-mg=\frac{36\times 13.09^2}{3.5}+36\times 9.8=2115.23 N$

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Ayoung kid of mass m = 36 kg is swinging on a swing. the length from the top of the swing set to the...

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