Question

The passengers in a roller coaster car feel 50% heavier than their true weight as the car goes through a dip with a 20 m radius of curvature.What is the car's speed at the bottom of the dip?

Answer 1

Answer:

v= 10 m/s

Explanation:

 Given that

Radius ,r= 20 m

The total wight R

$R=W+\dfrac{W}{2}$      (  50% heavier)

Lets take ,mass = m kg

$R=mg+\dfrac{mg}{2}$

Now by applying Newton's Second law

Total Force$F= mg+\dfrac{mv^2}{r}$

v=speed of the car at the bottom

Now by balancing the above forces

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

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

$\dfrac{g}{2}= \dfrac{v^2}{r}$

$v=\sqrt{\dfrac{gr}{2}}$

$v=\sqrt{\dfrac{10\times 20}{2}}\ m/s$             ( take g= 10 m/s²)

v= 10 m/s

Answer 2

Answer:

Explanation:

Force = weight + 50% of weight = 1.5 mg

Weight = mg

Let v be the speed.

radius, r = 20 m

According to the Newtons second law

F = mg + mv²/r

1.5 mg = mg + mv²/r

0.5 g = v²/r

v²  = 0.5 x 9.8 x 20

v = 9.8 m/s