Question #114453

A 48.0 kg in-line skater begins with a speed of 2.2 m/s. Friction also does (-150) J of work on her. Assume that se did not push on the ground any more. If her final speed is 5.9 m/s,


a) determine the change (final-initial) in her gravitational potential energy.


b) By how much, and in which direction (up or down), has her weight changed?


Please answer using grade 11 Conservation of Energy and Momentum .



1
Expert's answer
2020-05-08T16:25:52-0400

As per the given question,

mass of the line skater (m)=48kg(m)=48 kg

speed of the line skater (u)=2.2m/sec(u)=2.2 m/sec

work done by friction force (w)=150J(w)=-150J

her final speed (v)=5.0m/sec(v)=5.0m/sec

a) Applying the conservation of energy,

Initial potential energy (U1)=mu22(U_1)=\dfrac{mu^2}{2}

Final energy (U2)=mv22150(U_2)=\dfrac{mv^2}{2}-150

so, change in PE=48×5.92248×2.222150=(835.44232.32150)J=453.12J= \dfrac{48\times 5.9^2}{2}-\dfrac{48\times 2.2^2}{2}-150=(835.44-232.32-150 )J=453.12J


b)Weight of the skater is working downwards, because as she is moving downwards, her speed is getting increase, it means it is getting accelerate towards the ground.


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