Question

A student of mass M = 78 kg takes a ride in a frictionless loop-the-loop at an amusement park. The radius of the loop-the-loop is R = 20 m. The force due to the seat on the student at the top of the loop-the-loop is FN = 364 N and is vertically down. What is the apparent weight of the student at the bottom of the loop-the-loop?

Answer 1

Answer:

Explanation:

given,

mass M = 78 kg

radius of loop R = 20 m

F n = 364 N

now using newton's second law

$\dfrac{mu^2}{r}= N + mg$

$\dfrac{78\ u^2}{20}= 364 + 78\times 9.8$

$u^2= \dfrac{(364 + 78\times 9.8)\times 20}{78}$

$u = \sqrt{289}$

u = 17 m/s

using equation of motion

v² = u² + 2 ah

h = 2 × radius of loop

h = 2 × 20 = 40 m

v² = 17² + 2 × 9.8 × 40

v = 32.76 m/s

Apparent weight

$W = mg +\dfrac{mv^2}{r}$

$W = 78\times 9.8 +\dfrac{78 \times 32.76^2}{20}$

W = 4949.94 N

hence, the apparent weight is W = 4949.94 N

Answer 2

The apparent weight of the student at the bottom of the loop-the-loop is mathematically given as

W = 4949.94 N

What is the apparent weight of the student at the bottom of the loop-the-loop?

Question Parameter(s):

A student of mass M = 78 kg

he radius of the loop-the-loop is R = 20 m.

The force due to the seat on the student at the top of the loop-the-loop is FN = 364 N

Generally, the equation for the Newtons motion law   is mathematically given as

$\frac{mu^2}{r}= N + mg$

Therefore

$\frac{78\ u^2}{20}= 364 + 78*9.8$

$u^2= \frac{(364 + 78*9.8)*20}{78}$

u = 17 m/s

With

v^2 = u^ + 2 ah

Hence

v^2 = 17^2 + 2 * 9.8 * 40

v = 32.76 m/s

In conclusion, The apparent weight is

$W = 78* 9.8 +\frac{78 * 32.76^2}{20}$

W = 4949.94 N

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