Answer:
[tex]\Delta s = 1.28\,m[/tex]
Explanation:
The situation can be analyzed by means of the Principle of Energy Conservation and Work-Energy Theorem. The non-conservative force is the friction between ice and skates.
[tex]K_{A} = K_{B} + W_{friction}[/tex]
[tex]W_{friction} = K_{A}-K_{B}[/tex]
[tex]F_{friction} \cdot \Delta s = K_{A} - K_{B}[/tex]
[tex]\Delta s = \frac{K_{A}-K_{B}}{F_{friction}}[/tex]
[tex]\Delta s = \frac{\frac{1}{2}\cdot m \cdot (v_{A}^{2}-v_{B}^{2}) }{0.25\cdot m \cdot g}[/tex]
[tex]\Delta s = \frac{\frac{1}{2} \cdot (v_{A}^{2}-v_{B}^{2}) }{0.25 \cdot g}[/tex]
[tex]\Delta s = \frac{\frac{1}{2}\cdot [(3\,\frac{m}{s} )^{2}-(1.65\,\frac{m}{s} )^{2}] }{0.25\cdot (9.807\,\frac{m}{s^{2}} )}[/tex]
[tex]\Delta s = 1.28\,m[/tex]
Answer:
Explanation:
Given:
Initial velocity, vi = 3 m/s
Final velocity, vf = 1.65 m/s
Friction force, Fr = 25% of mass, m × acceleration, g
Using work-energy theorem,
Kinetic energy, Ek = - workdone by friction
Ek = 1/2 × m × (vf^2- vi^2)
1/2 × m × (vf^2- vi^2) = - u × m × g × d
u × m × g = 0.25 × m × g;
1/2 × m × (1.65^2 - 3^2) = - 0.25 × m × g × d
Solving for length of path = distance, d = (1.65^2 - 3^2)/-(0.5 × 9.8)
= 1.28 m.
Distance = 1.28 m
