Question

An unsecured 15 kg steel toolbox slides down the back of a Ute at an angle of 24-degrees to the horizontal. If the coefficient of kinetic friction for steel on steel is k = 0.57, what is the acceleration of the toolbox during this motion? Select one: O A. 3.47 m/s2 O B. 4.34 m/s2 O C. 1.12 m/s2 O D. 2.13 m/s2

Answer 1

1. The above problem is related to motion on inclined surfaces.
2. All the steps are clearly mentioned and easy to understand.
3. Hope the answer will help you. Thank you for your support.
4. 
   R imgsing to 0 24° ingcoso 7 Solution 1 Given, a steel toolbox having mass is m = 15 kg slides down the back of a ute. The coefficient of kinetic friction for steel on ang steel M = 0.57 So from figure The reaction force R = mga 15% 9.81 x Cool 24° 17 and Let the acceleration of the toolbox 'a' Therefore fama = my simo f I ng sino Mu RJ ma = | 158_9.8 LX 3in24° 0.57 X45*9.89 X 06 24° ma = |-16-771 cos ma A A G 16.77 = 4.12 m/s? 15 Therefore the answer will be option c.