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svetoff [14.1K]

2 years ago

8

A boy in a wheelchair (total mass 54.5 kg) 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 6.60 m/s. Assume 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.

1 answer:

babymother [125]2 years ago

3 0

Answer:

630.75 j

Explanation:

from the question we have the following

total mass (m) = 54.5 kg

initial speed (Vi) = 1.4 m/s

final speed (Vf) = 6.6 m/s

frictional force (FF) = 41 N

height of slope (h) = 2.1 m

length of slope (d) = 12.4 m

acceleration due to gravity (g) = 9.8 m/s^2

work done (wd) = ?

we can calculate the work done by the boy in pushing the chair using the law of law of conservation of energy

wd + mgh = (0.5 mVf^2) - (0.5 mVi^2) + (FF x d)

wd = (0.5 mVf^2) - (0.5 mVi^2) + (FF x d) - (mgh)

where wd = work done

m = mass

h = height

g = acceleration due to gravity

FF = frictional force

d = distance

Vf and Vi = final and initial velocity

wd = (0.5 x 54.5 x 6.9^2) - (0.5 x 54.5 x 1.4^2) + (41 x 12.4) - (54.5 X 9.8 X 2.1)

wd = 630.75 j

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$\displaystyle T=48.86\ N$

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<u>Net Force </u>

The second Newton's law explains how to understand the dynamics of a system where several forces are acting. The forces are vectorial magnitudes which means the x and y coordinates must be treated separately. For each component, the net force must equal the mass by the acceleration, i.e.

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Operating and rearranging

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Solving for T:

$\displaystyle T=\frac{a+\mu\ g}{cos\theta +\mu\ sin\theta }\ m$

Plugging in the given values:

$\displaystyle T=\frac{0.4+0.2(9.8)}{cos30^o+0.2\ sin30^o }\ .20$

$\boxed{\displaystyle T=48.86\ N}$

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