Question

A box of mass 17.6 kg with an initial velocity of 2.25 m/s slides down a plane, inclined at 19◦ with respect to the horizontal. The coefficient of kinetic friction is 0.48. The box stops after sliding a distance x. 17.6 kg µk = 0.48 2.25 m/s 19◦

How far does the box slide? The acceleration due to gravity is 9.8 m/s 2 . The positive x-direction is down the plane. Answer in units of m.

Answer 1

Answer:

2.01 m

Explanation:

Given parameters:

Mass, m = 17.6 kg

Velocity, v = 2.25 m/s

Degree = 19°

Coefficient of kinetic friction, $\mu_k$ = 0.48

To find the distance, we need to use Work-Energy principle.

So, total energy must be equal to the work done by friction. Here total energy is sum of the initial kinetic energy and the loss in potential energy.

$E_K = \frac{1}{2}mv^2\\E_P = mgxsin(\theta)\\W_{fr} = \mu_k mgcos(\theta)x$

So, $\frac{1}{2}mv^2 + mgxsin(\theta) = \mu_k mgcos(\theta)x$

$x = \frac{\frac{1}{2}mv^2}{\mu_k mgcos(\theta)-mgsin(\theta)} = \frac{\frac{1}{2}* 2.25^2}{0.48*9.8*cos(19)-9.8*sin(19)} = 2.01$ m