Question

1. A child slide down an inclined plane of length 10 m at an angle of 45°. If the coefficient friction between the child and the plane is 0.1, evaluate The velocity just before touching the bottom of the plane.
​

Answer 1

Answer:

The speed at the bottom is 11.2 m/s.

Explanation:

length, s = 10 m

Angle, A = 45 degree

coefficient of friction = 0.1

let the velocity is v.

The acceleration is given by

$a = g sin A - \mu g cos A \\\\a = 9.8 (sin 45 - 0.1 cos 45)\\\\a = 6.24 m/s^2$

Use third equation of motion

$v^2 = u^2 + 2 a s \\\\v^2 = 0 + 2 \times 6.24 \times 10 \\\\v = 11.2 m/s$