1. *A car is going over the top of a hill whose curvature approximates a circle of radius 200 m. At what velocity will the occupants of the car appear to weigh 20% less than their normal weight (or
their normal weight times 0.8)?

The velocity is 19.81 m/sec at which the occupants of the car appear to weigh 20% less than their normal weight and this can be determined by using the formula of centripetal force.

Given :

A car is going over the top of a hill whose curvature approximates a circle of radius 200 m.

The centripetal force acting on the car is given by the formula:

  ---- (1)

where m is the mass of a car, v is the velocity of the car and r is the radius of curvature.

Now, it is given that the occupants of the car appear to weigh 20% less than their normal weight that means:

  --- (2)

Now, equate equation (1) with equation (2).

v = 19.81 m/sec

The velocity is 19.81 m/sec at which the occupants of the car appear to weigh 20% less than their normal weight.

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The velocity of motion at which the occupants of the car appear to weigh 20% less than their normal weight is approximately 19.81 m/s

Explanation:

The given parameters are;

The curvature of the hill, r = 200 m

Due to the velocity, v, the occupants weight = 20% less than the normal weight

The outward force of an object due to centripetal (motion) force is given by the following equation;

Where;

r = The radius of curvature of the hill = 200 m

Given that the weight of the occupants, W = m × g, we have;

v = √(0.2 × g × r)

By substitution, we have;

v = √(0.2 × 9.81 × 200) ≈ 19.81

The velocity of motion at which the occupants of the car appear to weigh 20% less than their normal weight ≈ 19.81 m/s.