Question

question 14 (1 point)
a 10-kg mass slides down a flat hill that makes an angle of 10° with the horizontal. if friction is negligible, what is the resultant force on the sled?
1.7 n
17 n
97 n
98 n

Explanation:

Step1: Identify forces on the mass

The mass is on an inclined plane (hill) with negligible friction. The forces acting on it are gravity (\(mg\)) and the normal force. The resultant force along the incline is due to the component of gravity parallel to the incline.

Step2: Calculate gravitational force

First, find the weight of the mass: \(W = mg\), where \(m = 10\,\text{kg}\) and \(g = 9.8\,\text{m/s}^2\). So \(W = 10 \times 9.8 = 98\,\text{N}\) (this is the force due to gravity, acting vertically downward).

Step3: Find component parallel to incline

The component of gravity parallel to the incline (which is the resultant force here, since friction is negligible) is given by \(F_{\text{parallel}} = mg\sin\theta\), where \(\theta = 10^\circ\).

Substitute the values: \(F_{\text{parallel}} = 10 \times 9.8 \times \sin(10^\circ)\).

We know that \(\sin(10^\circ) \approx 0.1736\).

So \(F_{\text{parallel}} = 10 \times 9.8 \times 0.1736 \approx 17\,\text{N}\) (after calculation: \(10\times9.8 = 98\); \(98\times0.1736 \approx 17\)).

Answer:

17 N (corresponding to the option "17 N")