A luge and its rider, with a total mass of 84 kg, emerge from a downhill track onto a horizontal straight track with an initial speed of 41 m/s. Assume that they stop with a constant deceleration of 3.5 m/s2.
(a) What magnitude F is required for the decelerating force?
(b) What distance d do they travel while decelerating?
(c) What work W is done on them by the decelerating force?
(d) What is F for a deceleration of 7.0 m/s2?
(e) What is d for a deceleration of 7.0 m/s2?
(f) What is W for a deceleration of 7.0 m/s2?
F= ma
Vf^2=Vi^2 + 2ad solve for d.
Work = F*d
a) F = ma
|F| = 84kg*-3.5m/s^2
b)v=0
x= -(41m/s)^2/(2(-3.5m/s^2))
c)W is going in the -x direction because it's decelerating
W= -Fd
Then you just apply these equations to d,e, and f.
To solve this problem, we can use the equations of motion for constant acceleration. Let's solve the problem step by step.
(a) What magnitude F is required for the decelerating force?
The force required for deceleration can be found using Newton's second law, which states that force is equal to mass multiplied by acceleration:
F = m * a
Here, the mass (m) is the total mass of the luge and the rider, which is given as 84 kg. The acceleration (a) is the constant deceleration of 3.5 m/s^2.
F = 84 kg * 3.5 m/s^2
F = 294 N
Therefore, the magnitude of the decelerating force required is 294 N.
(b) What distance d do they travel while decelerating?
To find the distance traveled during deceleration, we can use the following equation of motion:
v^2 = u^2 + 2as
where:
v = final velocity (0 m/s, as they stop)
u = initial velocity (41 m/s)
a = acceleration (-3.5 m/s^2)
s = distance traveled
Rearranging the equation, we get:
s = (v^2 - u^2) / (2a)
Substituting the known values:
s = (0^2 - 41^2) / (2*(-3.5))
s = -1681 / -7
s = 240.14 m (rounded to two decimal places)
Therefore, the distance they travel while decelerating is 240.14 meters.
(c) What work W is done on them by the decelerating force?
The work done on an object can be calculated using the equation:
W = F * d * cosθ
Here, F is the force (294 N) and d is the displacement distance (240.14 m). Since the force is opposing the motion, the angle between the force and displacement is 180 degrees, so cosθ is -1.
W = 294 N * 240.14 m * (-1)
W = -70,541.16 J (rounded to two decimal places)
Therefore, the work done on them by the decelerating force is -70,541.16 J.
(d) What is F for a deceleration of 7.0 m/s^2?
Using the same equation as in part (a), we can calculate the force:
F = m * a
Substituting the given values:
F = 84 kg * 7.0 m/s^2
F = 588 N
Therefore, the magnitude of the decelerating force required for an acceleration of 7.0 m/s^2 is 588 N.
(e) What is d for a deceleration of 7.0 m/s^2?
Using the equation of motion as in part (b), with v = 0, u = 41 m/s, and a = -7.0 m/s^2:
s = (0^2 - 41^2) / (2*(-7.0))
s = -1681 / -14
s = 120.07 m (rounded to two decimal places)
Therefore, the distance they travel while decelerating with an acceleration of 7.0 m/s^2 is 120.07 meters.
(f) What is W for a deceleration of 7.0 m/s^2?
Using the equation for work as in part (c), with F = 588 N and d = 120.07 m:
W = 588 N * 120.07 m * (-1)
W = -70,562.16 J (rounded to two decimal places)
Therefore, the work done on them by the decelerating force for an acceleration of 7.0 m/s^2 is -70,562.16 J.