12. A person in a wheelchair did 70.8 J of work on the chair to get it moving from rest at the bottom of a ramp, taking 1.00 s to do so. She turned the wheel (which has a radius of 40.0 cm) one quarter turn before letting go and rolling back down. If the combined mass of her and the chair was 98.0 kg, what was her final velocity when she passed back over the bottom of the ramp?

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WILL GIVE

Her final velocity at she moves over the the bottom of the ramp is approximately 1.20204 m/s

Explanation:

The data of the energy used by the person in the wheel chair are;

The work done by the person, W = 70.8 J

The initial position of the wheel chair = At the bottom of the ramp

The time the person takes to do the work, t = 1.00 s

The number of turns she turned the wheel chair = 1/4 turns

The radius of the wheel of the wheel chair, r = 40 cm

The mass of her and the wheel chair, m = 98.0 kg

By the conservation of energy principle, we have;

The total mechanical energy of the motion, M.E. = The amount of work done by her in moving up the ramp = 70.8 J

M.E. = P.E. + K.E. = Constant

Where;

P.E. = Potential energy

K.E. = Kinetic energy

At the height she reached the ramp, M.E. = P.E. = 70.8 J

At the bottom of the ramp, P.E. = 0 J, therefore;

K.E. = 70.8 J

We have;

K.E. = (1/2)·m·v²

Where;

m = The mass of her and the wheelchair = 98.0 Kg

v = Her final velocity at the bottom of the ramp

By substitution, we have;

70.8 J =  (1/2) × 98.0 Kg × v²

v² = 70.8/((1/2) × 98.0) = 354/245

v = √(1,770)/35 ≈ 1.20204

Her final velocity as she passed back over the bottom of the ramp, v ≈ 1.20204 m/s.

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