A skateboarder traveling at 4.45 m/s can be stopped by a strong force in 1.82 s and by a weak force in 5.34 s. The mass of the skateboarder and the skateboard is 53.6 kg. What is the impulse on the skateboarder? What is the average force on the skateboarder in each of these stops? Analyze and Sketch the Problem

List Knowns and Unknowns

Knowns Unknowns

m = ∆tstrong force = F∆t =

∆vi = ∆tweak force = Fstrong force =

∆vf = Fweak force =

SOLVE FOR THE UNKNOWNS
a. Determine the initial and final momentum

b. Apply the impulse-momentum theorem to determine the impulse.

c. Use the impulse to determine the force needed to stop the skateboarder

EVALUATE THE ANSWER
Explain why the magnitude of the force is realistic considering the times

Explanation:

The relation between momentum "P" and impulse "I " is given by the impulse momentum theorem given below,

i.e. impulse is equal to change in momentum.

I= ∆P= P₂ - P₁

F∆t=mv₂ - mv₁

here m is the mass of skateboarder= m= 53.6 kg

initial velocity v₁ = 4.45m/s

final velocity v₂ = 0

now impulse is

I =0 - 53.6 x 4.45 = -235.52 kgm/s

here the final momentum is zero as the final velocity is zero and the initial momentum P₁= mv₁ = 235.52kgm/s

now the strong force that is applied for 1.82 seconds can be calculated as

F= - I / t

F = - 235.52/ 1.82 = - 131 N

in second step the weak force that is applied for 5.34 s can be calculated as

F = - I / t = - 235.52 / 5.34 = - 44.6 N

the negative sign of impulse and applied force indicates that the force is opposite to the direction of motion of the skateboarder.

d eras and periods

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