Downhill Racing Wheelchairs
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Downhill Racing Wheelchairs
Energy question, please help me ?
A boy in a wheelchair (total mass, 48.1 kg) wins a race with a skateboarder. He has a speed of 1.38 m/s at the crest of a slope 2.37 m high and 12.8 m long. At the bottom of the slope, his speed is 6.18 m/s. If air resistance and rolling resistance can be modeled as a constant frictional force of 42.8 N, calculate the work he did in pushing forward on his wheels during the downhill ride.
I got the answer : -792.28 J
i said w=(m/2)(vf^2-vi^2)-mgh- ff*12.8
please help ..
His translational kinetic energy at the top of the hill is
KE1 = 1/2 mv^2 = (48.1 kg)*(1.38 m/s)^2
and at the bottom of the hill is
KE2 = 1/2 mv^2 = (48.1 kg)*(6.18 m/s)^2
His gravitational potential energy has decreased by
PE = mgh = (48.1 kg)*(9.8 m/s^2)*(2.37 m)
and the energy he's lost to friction is
Wf = F x d = (42.8 N)*(12.8 m)
If W is the work the boy does, then
KE2 - KE1 = PE + W - Wf
W = KE2 - KE1 - PE - Wf
Looks very much like your expression, doesn't it. The arithmetic yields the same result as yours, too. In terms of arithmetic, you've got the right answer.
The problem itself though is very, very broken. A wheelchair has very big wheels, and if you've ever ridden a bicycle than you know that those rotating wheels will have soaked up a lot of energy. However, there's not enough information to calculate moment of inertia of the wheels or their angular velocity, so the problem assumes that rotational KE isn't relevant to the solution. That's really wrong. You though are really right, so congratulations - you've correctly solved a problem that doesn't correspond to any reality that you've ever known.
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A Death Among Heroes - Downhill Wheelchair Racing
physics geniuses, can u figure this out??
A boy in a wheelchair (total mass 56.5 kg) wins a race with a skateboarder. He has speed 1.40 m/s at the crest of a slope 2.10 m high and 12.4 m long. At the bottom of the slope his speed is 6.50 m/s. If air resistance and rolling resistance can be modeled as a constant friction force of 41.0 N, find the work he did in pushing forward on his wheels during the downhill ride.
I realize work is equal to change in kinetic energy but how do i figure that out. i tried 1/2mv^2 just isnt working. can u help please?
1. work= scalar multiplication of vectors(F and s)
where s is distance
W=|F|*|s|*cos(a)
aleph = arctg(2.10/12.4) and is the slope angle
s=sqrt(2.1^2+12.4^2)=12.58m
Ff=friction force.
|Fg|=m*g=9.81*56.5=554.27N gravimetric force
|Fn|= - m*g*cos(aleph) normal force (is perpendicular to the skew) no need to compute it.
Fv=Fn+Fg=m*g*sin(aleph) = 554.27*(2.10/12.58) = 92.5N (dont forget Fn and Fg form angle 180-aleph)
Wg=s*Fv=12.58*92.5=1164J where Wg is the work of gravimetric force
Wf=Ff*s= -41.0*12.58=-515,78J is the work of friction force and is negative
v, ...initial speed 1.40 m/s
v,, .... resulting 6.50 m/s
Wt=1/2m*(v,,)^2-1/2m*(v,)^2 =1138,19J is the total work
Wt=Wg+Wb-Wf
Wb=Wt-Wg+Wf=1138.19-1164+515.78=490J
490J is work the boy did. (You have to add Wf cause he had to overcame the friction but the gravity helped him)
Also you can compute it by taking:
Wg=m*g*h
1164J=9.81*56.5*2.1=1163.967J
without rounding it holds
Wt-m*g*h=Wf=Wb
This is the last and final edit.
