Question

59. (II) The crate shown in Fig. 4-60 lies on a plane tilted at an angle A = 25.0° to the horizontal, with Mk 0.19. (a) Determine the acceleration of the crate as it slides down the plane. (b) If the crate starts from rest 8.15 m up along the plane from its base, what will be the crate's speed when it reaches the bottom of the incline?​

Answer 1

Explanation:

a) We need to write down first Newton's 2nd law as applied to the given system. The equations of motion for the x- and y-axes can be written as follows:

$x:\;\;\;\;\;mg\sin 25° - \mu_kN = ma\;\;\;\;\;\;(1)$

$y:\;\;\;\;\;N - mg\cos 25° = 0\;\;\;\;\;\;\;\;\;(2)$

From Eqn(2), we see that

$N = mg\cos 25°\;\;\;\;\;\;\;(3)$

so using Eqn(3) on Eqn(1), we get

$mg\sin 25° - \mu_kmg\cos 25° = ma$

Solving for the acceleration, we see that

$a = g(\sin 25° - \mu_k\cos 25°)$

$\;\;\;\;= 2.45\:\text{m/s}^2$

b) Now that we have the acceleration, we can now solve for the velocity of the crate at the bottom of the plane. Using the equation

$v^2 = v_0^2 + 2ax$

Since the crate started from rest, $v_0 = 0.$ Thus our equation reduces to

$v^2 = 2ax \Rightarrow v = \sqrt{2ax}$

$v = \sqrt{2(2.45\:\text{m/s}^2)(8.15\:\text{m})}$

$\;\;\;\;= 6.32\:\text{m/s}$