Question

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?

Answer 1

Part a)

`F = 205 N`

Part b)

`d = 109.7 m`

Part c)

`W = -22500 J`

Part d)

`F = 410 N`

Part e)

`d = 54.85 m`

Part f)

`W = -22500 J`

Step-by-step 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

`F = (50)(4.1)`

`F = 205 N`

Part b)

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

then we have

`v_f^2 - v_i^2 = 2 a d`

`0 - 30^2 = 2(-4.1) d`

`d = 109.7 m`

Part c)

Work done to stop the rider

`W = F .d`

`W = F d cos 180`

`W = 205 (109.7) cos180`

`W = -22500 J`

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

`F = (50)(8.2)`

`F = 410 N`

Part e)

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

then we have

`v_f^2 - v_i^2 = 2 a d`

`0 - 30^2 = 2(-8.2) d`

`d = 54.85 m`

Part f)

Work done to stop the rider

`W = F .d`

`W = F d cos 180`

`W = 410 (54.85) cos180`

`W = -22500 J`