Question

4. Two people with the same height are doing drop landings (where you step off of a box onto the floor) off a box that is 40 cm high. The first person weighs 600 N. The second person weighs 700 N.

a. Using Newton’s laws, what is the velocity of each person when the first land on the floor? It takes 0.285 seconds for the people to go from the box to the floor .

b. If the first person takes 0.4 seconds to come to rest and the second person takes 0.45 seconds to come to rest, which person experiences the larger average force from the floor? Explain your answer with data.

Answer 1

(a) You need tobuse kinematic equations,

Initial speed = 0 m/s (they drop)

g is acceleration due to gravity 9.81 m/s2 downward

40 cm = 0.4 m

s = ut + (1/2)gt²

=> 0.4 = 0*t + (1/2)*(9.81)*(t²)

=> t = 0.286 s

a = (v-u)/t

=> 9.81 = (v-0)/0.286

=> v = 2.8 m/s

Since, the persons are free falling, mass has no effect and both will land on the floor at a velocity of 2.8 m/s in 0.286s

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b) Impulse mometum theroem says that the impulse which is the product of average force × time = change in momentum which is equal to product of mass × change in velocity

Impulse = change on momentum

=> F*t = m(v-u)

For the first person,

m = 600/g = 600/9.81 = 61.16 kg

F₁*0.4 = (61.16)(0 - 2.8)

=> F₁ = -428.12 N (negative sign indicates that the force is in upwards direction)

For the second person,

m = 600/g = 700/9.81 = 71.36 kg

F₂*0.45 = (71.36)(0 - 2.8)

=> F₂ = -444.02 N (negative sign indicates that the force is in upwards direction)

Therefore, the second person experiences larger average force from the floor since his weight is more