Answer to Question #135243 in Physics for 5465JUY

Question

1. A block slides down a frictionless plane having an inclination of = 15.0 0 .The block starts from rest at the top, and the length of the incline is 2.00 m.

Find;

(a) the acceleration of the block

(b) its speed when it reaches the bottom of the incline. 

Answer 1

a) There are two forces that act on the block: the force of gravity (or the weight) $mg$ and the normal force $N$. The force of gravity can be resolved into two perpendicular components ($F_{\shortparallel} = mgsin\theta$ and $F_{\bot} = mgcos\theta$).

We can find the acceleration of the block from the Newton's Second Law of Motion:

$$\sum F_x = ma_x,$$ $$mgsin\theta = ma,$$ $$a = gsin\theta = 9.8 \ \dfrac{m}{s^2} \cdot sin15^{\circ} = 2.54 \ \dfrac{m}{s^2}.$$

b) We can find the speed of the block when it reaches the bottom of the incline from the kinematic equation:

$$v_f^2 - v_i^2 = 2as,$$

here, $v_i = 0$ is the initial speed of the block, $v_f$ is the final speed of the block when it reaches the bottom of the incline, $a$ is the acceleration of the block, $s = 2.0 \ m$ is the distance traveled by the block.

Then, from this formula we can calculate the speed of the block when it reaches the bottom of the incline:

$$v_f = \sqrt{2as} = \sqrt{2 \cdot 2.54 \ \dfrac{m}{s^2} \cdot 2.0 \ m} = 3.19 \ \dfrac{m}{s}.$$

Answer:

a) $a = 2.54 \ \dfrac{m}{s^2}.$

b) $v_f = 3.19 \ \dfrac{m}{s}.$