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bogdanovich [222]
3 years ago
9

Holding onto a tow rope moving parallel to a frictionless ski slope, a 68.7 kg skier is pulled up the slope, which is at an angl

e of 6.7° with the horizontal. What is the magnitude Frope of the force on the skier from the rope when (a) the magnitude v of the skier's velocity is constant at 1.90 m/s and (b) v = 1.90 m/s as v increases at a rate of 0.150 m/s2?
Physics
1 answer:
Furkat [3]3 years ago
3 0

Answer:

a) F = 78.606\,N, b) F = 88.911\,N

Explanation:

a) Let consider two equations of equilibrium, the first parallel to ski slope and the second perpendicular to that. The equations are, respectively:

\Sigma F_{x'} = F - m\cdot g \cdot \sin \theta = 0\\\Sigma F_{y'} = N - m\cdot g \cdot \cos \theta = 0

The force on the skier is:

F = m \cdot g \cdot \sin \theta

F = (68.7\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot \sin 6.7^{\textdegree}

F = 78.606\,N

b) The equations of equilibrium are the following:

\Sigma F_{x'} = F - m\cdot g \cdot \sin \theta = m\cdot a\\\Sigma F_{y'} = N - m\cdot g \cdot \cos \theta = 0

The force on the skier is:

F = m\cdot (a + g \cdot \sin \theta)

F = (68.7\,kg)\cdot (0.150\,\frac{m}{s^{2}}+9.807\,\frac{m}{s^{2}}\cdot \sin 6.7^{\textdegree})

F = 88.911\,N

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