Questions: Uno sciatore scende con velocità costante di 10 m / s lungo un pendio inclinato di 30° rispetto all'orizzontale. La sua massa è 70 kg. Trascura l'attrito con l'aria. Calcola il lavoro compiuto dalla forza d'attrito con il suolo in 1,0 s. [-3,4 x 10^3 J]

Solution Steps

The skier is moving with a constant velocity of \(10 \, \mathrm{m/s}\). In \(1.0 \, \mathrm{s}\), the distance traveled \(d\) is: \[ d = v \cdot t = 10 \, \mathrm{m/s} \cdot 1.0 \, \mathrm{s} = 10 \, \mathrm{m} \]

The gravitational force component along the incline \(F_{\parallel}\) is given by: \[ F_{\parallel} = mg \sin \theta \] where \(m = 70 \, \mathrm{kg}\), \(g = 9.81 \, \mathrm{m/s^2}\), and \(\theta = 30^\circ\). Substituting the values: \[ F_{\parallel} = 70 \, \mathrm{kg} \cdot 9.81 \, \mathrm{m/s^2} \cdot \sin 30^\circ = 70 \cdot 9.81 \cdot 0.5 = 343.35 \, \mathrm{N} \]

Since the skier is moving with a constant velocity, the net force along the incline is zero. This means the friction force \(F_f\) must be equal and opposite to the component of the gravitational force along the incline: \[ F_f = -F_{\parallel} = -343.35 \, \mathrm{N} \] The work done by the friction force \(W_f\) over the distance \(d\) is: \[ W_f = F_f \cdot d = -343.35 \, \mathrm{N} \cdot 10 \, \mathrm{m} = -3433.5 \, \mathrm{J} \]

Final Answer