JEE Main Physics question
The coefficient of friction between a body and the surface of an inclined plane at 45° is 0.5. If g = 9.8 m/s², the acceleration of the body downwards in m/s² is
Answer: (A)
Solution
a = g(sinθ − μcosθ) = 9.8(sin45° − 0.5cos45°) = 9.8×cos45°×(1−0.5) =
See the full step-by-step solution
Create a free account to read the complete working, and practise questions like this in a timed test.
Create a free account