2017 ?International Practice Exam MCQ

2017 ?International Practice Exam MCQ

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A skier is traveling down a slope that is inclined 30° ?above the horizontal. The coefficient of kinetic friction between the skier and the slope is 0.10 . ?Which of the following best describes the acceleration of the skier?

(A) ?It is zero.

(B) ?It is about 4.0ms2.

(C) ?It is about 9.0ms2.

(D) ?It is about 10ms2.

(E) ?It cannot be determined without knowing the mass of the skier.

The correct answer and explanation is:

The correct answer is (B) It is about 4.0 m/s².

Explanation:

To find the acceleration of the skier, we need to analyze the forces acting on the skier as they move down the slope. The skier is affected by the following forces:

1. Gravitational Force (Weight): This is the force due to the skier’s mass and gravity, acting vertically downward.
2. Normal Force: The contact force between the skier and the slope, perpendicular to the surface of the slope.
3. Frictional Force: The force opposing the skier’s motion, caused by the interaction between the skier and the slope. It is proportional to the normal force and is given by ffriction=μkNf_{\text{friction}} = \mu_k N, where μk\mu_k is the coefficient of kinetic friction and NN is the normal force.

Steps to Calculate the Acceleration:

1. Decompose the Gravitational Force: The weight of the skier can be decomposed into two components:
   • The component parallel to the slope: F∥=mgsin⁡θF_{\parallel} = mg \sin \theta
   • The component perpendicular to the slope: F⊥=mgcos⁡θF_{\perp} = mg \cos \theta where:
   • mm is the skier’s mass
   • gg is the acceleration due to gravity (9.8 m/s²)
   • θ\theta is the angle of the slope (30°).
2. Normal Force: The normal force is equal to the perpendicular component of the gravitational force: N=mgcos⁡θN = mg \cos \theta.
3. Frictional Force: The frictional force is ffriction=μkN=μkmgcos⁡θf_{\text{friction}} = \mu_k N = \mu_k mg \cos \theta.
4. Net Force: The net force in the direction of the slope is the difference between the parallel component of gravity and the frictional force: Fnet=mgsin⁡θ−μkmgcos⁡θF_{\text{net}} = mg \sin \theta – \mu_k mg \cos \theta
5. Acceleration: Using Newton’s second law (F=maF = ma), we can find the acceleration: a=Fnetm=g(sin⁡θ−μkcos⁡θ)a = \frac{F_{\text{net}}}{m} = g (\sin \theta – \mu_k \cos \theta) Substituting the given values:
   • g=9.8 m/s2g = 9.8 \, \text{m/s}^2
   • θ=30∘\theta = 30^\circ
   • μk=0.10\mu_k = 0.10
   a=9.8×(sin⁡30∘−0.10cos⁡30∘)a = 9.8 \times (\sin 30^\circ – 0.10 \cos 30^\circ) a=9.8×(0.5−0.10×0.866)a = 9.8 \times (0.5 – 0.10 \times 0.866) a=9.8×(0.5−0.0866)a = 9.8 \times (0.5 – 0.0866) a=9.8×0.4134≈4.0 m/s2a = 9.8 \times 0.4134 \approx 4.0 \, \text{m/s}^2

Thus, the skier’s acceleration is approximately 4.0 m/s², making (B) the correct choice.