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A luge and its rider, with a total mass of 50 kg, emerge from a downhill track onto a horizontal straight track with an initial speed of 30 m/s. If a force slows them to a stop at a constant rate of 4.1 m/s2, (a) what magnitude F is required for the force, (b) what distance d do they travel while slowing, and (c) what work W is done on them by the force? What are (d) F, (e) d, and (f) W if they, instead, slow at 8.2 m/s2?

Respuesta :

Answer:

Part a)

[tex]F = 205 N[/tex]

Part b)

[tex]d = 109.7 m[/tex]

Part c)

[tex]W = -22500 J[/tex]

Part d)

[tex]F = 410 N[/tex]

Part e)

[tex]d = 54.85 m[/tex]

Part f)

[tex]W = -22500 J[/tex]

Explanation:

Part a)

Mass of the rider = 50 kg

rate of deceleration of the object = 4.1 m/s/s

now we know by Newton's II law

F = m a

we have

[tex]F = (50)(4.1) [/tex]

[tex]F = 205 N[/tex]

Part b)

Since the rider is decelerated then let say it will be stopped after moving distance "d"

then we have

[tex]v_f^2 - v_i^2 = 2 a d[/tex]

[tex]0 - 30^2 = 2(-4.1) d[/tex]

[tex]d = 109.7 m[/tex]

Part c)

Work done to stop the rider

[tex]W = F .d [/tex]

[tex]W = F d cos 180[/tex]

[tex]W = 205 (109.7) cos180[/tex]

[tex]W = -22500 J[/tex]

Part d)

Mass of the rider = 50 kg

rate of deceleration of the object = 8.2 m/s/s

now we know by Newton's II law

F = m a

we have

[tex]F = (50)(8.2) [/tex]

[tex]F = 410 N[/tex]

Part e)

Since the rider is decelerated then let say it will be stopped after moving distance "d"

then we have

[tex]v_f^2 - v_i^2 = 2 a d[/tex]

[tex]0 - 30^2 = 2(-8.2) d[/tex]

[tex]d = 54.85 m[/tex]

Part f)

Work done to stop the rider

[tex]W = F .d [/tex]

[tex]W = F d cos 180[/tex]

[tex]W = 410 (54.85) cos180[/tex]

[tex]W = -22500 J[/tex]

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