Start learning →

A skier of total mass 60.0 kg is on a slope at 25 degrees to the horizontal with friction negligible. What is her acceleration down the slope?

A4.14 m/s^2
B8.88 m/s^2
C9.80 m/s^2
D0.42 m/s^2
Answer & Solution
Correct answer: A. 4.14 m/s^2
1. Choose axes with one parallel to the slope and one perpendicular to it. 2. There is no motion perpendicular to the slope, so only the parallel forces matter. 3. With friction negligible the only parallel force is the component of weight along the slope. 4. That component is the mass times the acceleration due to gravity times the sine of 25 degrees. 5. Newton's second law then gives the acceleration as the acceleration due to gravity times the sine of 25 degrees. 6. Substitute the values: (9.80 m/s^2)(0.4226) = 4.14 m/s^2. 7. The mass cancels, so every skier on that slope accelerates alike. 8. Quoting 9.80 m/s^2 forgets that only part of the weight acts along the slope. _Source: OpenStax College Physics (CC BY 4.0), Ch 4 "Dynamics: Force and Newton's Laws of Motion", section 4.5 Normal, Tension, and Other Examples of Forces_
Practise this question with Quill — JAMB UTME →

Keep practising Dynamics: Force and Newton's Laws of Motion

Browse all Dynamics: Force and Newton's Laws of Motion questions for JAMB UTME, or explore more Physics topics.

Start a Dynamics: Force and Newton's Laws of Motion lesson →
Related questions