A skier speeds down a smooth ski slope which is at an angle of θ = 23° with the horizontal. The mass of the skier is 71 kg. Take the downhill direction to be positive and uphill to be negative. (a) What net force (in N) is acting on the skier? (Indicate the direction with the sign of your answer.) N (b) What is the acceleration (in m/s2) experienced by the skier? (Indicate the direction with the sign of your answer.) m/s2 (c) How does the net force experienced by the skier change if the ski slope becomes steeper? increases decreases remains the same

Question

Priscilla Boyer

Physics

5 May, 04:58

A skier speeds down a smooth ski slope which is at an angle of θ = 23° with the horizontal. The mass of the skier is 71 kg. Take the downhill direction to be positive and uphill to be negative. (a) What net force (in N) is acting on the skier? (Indicate the direction with the sign of your answer.) N (b) What is the acceleration (in m/s2) experienced by the skier? (Indicate the direction with the sign of your answer.) m/s2 (c) How does the net force experienced by the skier change if the ski slope becomes steeper? increases decreases remains the same

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Answers (1)

Know the Answer?

Charlize Hubbard · Answer 1

(a) The net force is: 271.7 (i) N

(b) The acceleration is: 3.8 m/s²

(c) The net force increases

Explanation:

In order to answer the questions you have to apply Newton's Second Law:

∑F=mA

Where F is the force, m is the mass and A is the acceleration.

Applying it for the x-axis and y-axis (Where the x-axis is in the slope's direction and the y-axis is perpendicular to it)

x-axis: Wx = mA

y-axis: N-Wy=0 (because the velocity in the y-axis is constant)

Wx and Wy are the components of the weight vector.

You have to obtain the components of the weight. The slope is at an angle of 23° with the horizontal and the weight is perpendicular to the horizontal, therefore the angle formed by the weight and the slope is: 90-23=67°

Applying trigonometric identities for a right triangle to obtain the components of the weight:

Wx = mgCos (67°) = (71) (9.8) Cos (67°) = 271.9 N

Wy = mgSin (67°) = (71) (9.8) Sin (67°) = 640.5 N

Therefore the net force is the sum of the net force in x and the net force in y (which is zero)

F=Wx (i)

F = 271.7 N

Where i is the unit vector.

The acceleration is:

A=Wx/m=271.7/71 = 3.8 m/s²

For part (c):

If the ski slope becomes steeper it means that the angle increases. The net force is: mgCos (90-θ) which is equal to mgSin (θ). Therefore, if the angle increases then Sin (θ) increases and the force increases.