Question

application of forces
a child has a mass of 36 kg and is sitting on a seat on an amusement park ride. the ride makes the seat move up and down. determine the normal force acting on the child when
the child is
(a) moving up at a constant velocity of 12 m/s ans: 350 n up (b) moving down at a constant velocity of 14 m/s ans: 350 n up
(c) accelerating down at 1.8 m/s2

Explanation:

Step1: Recall Newton's Laws

For constant velocity (a and b), acceleration \(a = 0\). Use Newton's second law \(F_{net}=ma\). Forces on child: normal force \(F_N\) (up) and weight \(F_g = mg\) (down). So \(F_{net}=F_N - F_g=ma\). When \(a = 0\), \(F_N = F_g\). Calculate \(F_g=36\,\text{kg} \times 9.8\,\text{m/s}^2\approx 352.8\,\text{N}\approx 350\,\text{N}\) (up as normal force).

Step2: For part (c) (accelerating down)

Acceleration \(a = 1.8\,\text{m/s}^2\) (down). So \(F_{net}=F_g - F_N=ma\) (since net force is down). Rearrange: \(F_N = F_g - ma\). Calculate \(F_g = 36\times9.8 = 352.8\,\text{N}\), \(ma = 36\times1.8 = 64.8\,\text{N}\). Then \(F_N = 352.8 - 64.8 = 288\,\text{N}\) (up, as normal force direction is perpendicular to seat, upward).

Answer:

(a) \(350\,\text{N}\) [up], (b) \(350\,\text{N}\) [up], (c) \(288\,\text{N}\) [up]