Question

A 4.20–kg block is set into motion up an inclined plane with an initial speed of v_i = 7.40 m/s (see figure below). The block comes to rest after traveling d = 3.00 m along the plane, which is inclined at an angle of 𝜃 = 30.0° to the horizontal.

Answer 1

Answer

By energy conservation

So

$mgh+\frac{1}{2}mv^2=\mu mgcos\theta d$

$9.8(3.00sin30°)+\frac{1}{2}(7.40)^2=\mu (9.8)(cos30°)(3)$

So

$\mu=1.65$