Last month Dr. Hruda's electric bill showed that she used 249 kW-hr. How many Joules (kJ) did Dr. Hruda use?
1 answer:
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Answer:
71.76 m
Explanation:
We will solve this question using the work energy theorem.
The theorem explains that, the change in kinetic energy of a particle between two points is equal to the workdone in moving the particle from the one point to the other.
ΔK.E = W
In the attached free body diagram for the question, the forces acting on the puck are given.
ΔK.E = (final kinetic energy) - (initial kinetic energy)
Final kinetic energy = 0 J (since the puck comes to a stop)
Initial kinetic energy = (1/2)(m)(v²) = (1/2)(0.2)(26²) = 67.6 J
ΔK.E = 0 - 67.6 = - 67.6 J
W = Workdone between the starting and stopping points = (work done by the force of gravity) + (work done by frictional force)
Work done by the force of gravity = - mgh = - (0.2)(9.8)(h) = - 1.96 h
Workdone by the frictional force = F × d
F = μ N
μ = coefficient of kinetic friction = 0.30 (kinetic frictional force is the only frictional force that moves a distance of d, the static frictional force doesn't move any distance, so it does no work)
N = normal reaction of the plane surface on the puck = mg cos 30° = (0.2)(9.8)(0.866) = 1.697 N
F = μ N = 0.3 × 1.697 = 0.509 N
where d = distance along the incline that the puck travels.
d = h/sin 30° = 2h (from trigonometric relations)
Workdone by the frictional force = F × d = 0.509 × 2h = 1.02 h
ΔK.E = W = (work done by the force of gravity) + (work done by frictional force)
- 67.6 = - 1.96h + 1.02h
-0.942h = - 67.6
h = 71.76 m
