Question

Riders on the Tower of Doom, an amusement park ride, experience 2.0 s of free fall, after which they are slowed to a stop in 0.50 s.
A) What is a 65 kg rider's apparent weight as the ride is coming to rest?
B) By what factor does this exceed her actual weight?

Answer 1

Newton's second law and kinematics allow finding the answers for the changes in weight when the body stops are:

        A) The apparent weight is 3548 N directed upwards

      B)  The factor of change of weights is 5

Given parameters

• Free fall time t1 = 2.0 s

• The time tot stops at t2 = 0.5 s

• Mass m = 65 kg

To find

   A)  Apparent weight

   B)  Factor that exceeds the actual weight

Kinematics studies the movement of bodies, looking for relationships between the position, speed and acceleration of bodies.

This exercise we will do them in two parts

1st part. Freefall

We look for the velocity at the end of the trajectory, as part of rest the initial velocity is zero

            v = v₀ - g t

Where v is the velocity, vo the initial velocity, g the acceleration due to gravity and y time

            v = -gt

            v = -9.8 2.0

            v = -19.6 m / s

The negative sign indicates that the velocity is directed downwards

2nd part. Braking movement

We look for the acceleration to stop the body, in this case the final velocity is zero

            $v_f$ = v + a t

             0 = v + at

             a = $- \frac{v}{t}$- v / t

             a = $-\ \frac{-19.6}{0.5}$

             a = + 39.2 m / s²

The positive sign indicates that the acceleration is directed upward.

A) Newton's second law states that the force is proportional to the mass and the acceleration of the body

            ∑ F = m a

            F -W = m a

            F = m (a + g)

            F = 65 (39.2 + 9.8)

            F = 3185 N

Body weight is

             W = mg

             W = 65 9.8

             W = 637 N

therefore the apparent weight is

             ΔW = F -W

             ΔW = 3185 - 637

             Δw = 2548 N

B) the weight change factor

             Factor =$\frac{F}{W}$ F / W

             Factor =$\frac{3185}{937}$

             factor = 5

In conclusion, using Newton's second law and kinematics, we can find the answers for the changes in weight when the body stops are:

       A)  The apparent weight is 3548 N directed upwards

      B)  The factor of change of weights is 5

Learn more about Newton's second law and kinematics here:

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