Question

A box with mass 25.14 kg is sliding at rest from the top of the slope with height 13.30 m

and slope angle 30 degree, suppose the coefficient of friction of the slope surface is
0.25, find (neglect air resistance,take g=10 m/s^2)
The friction force experienced by the box.
00) The acceleration of the box along the slope.
(1) The time T required for the object to reach the bottom of the slope from the slope top.

Answer 1

Step-by-step explanation:

The first thing we are asked to find is the Force experienced by the box. That is found in the formula:

F - f = ma where F is the force exerted by the box, f is the friction opposing the box, m is the mass, and a is the acceleration (NOT the same as the pull of gravity). But F can be rewritten in terms of the angle of inclination also:

`wsin\theta-f=ma` where w is the weight of the box. We will use this version of the formula because it will help us answer the second question, which is to solve for a. Filling in:

First we need the weight of the box. Having the mass, we find the weight:

w = mg so

w = 25.14(10) so

w = 251.4 N (I am not paying any attention at all to the sig fig's here, since I noticed no one on this site does!) Now we have the weight. Filling that in:

251.4sin(30) - f = ma Before we go on to fill in for f, let's answer the first question. F = 251.4sin(30) so

F = 125.7 And in order to answer what a is equal to, we find f:

f = μ
`F_n` where Fn is the weight of the object.

f = .25(251.4) so

f = 62.85. Filling everything in now altogether to solve for a, the only missing value:

125.7 - 62.85 = 25.14a and

62.85 = 25.14a so

a = 2.5 m/s/s

Now we have to move on to another set of equations to answer the last part. The last part involves the y-dimension. In this dimension, what we know is that

a = -10 m/s/s

v₀ = 0 (it starts from rest)

Δx = -13.30 m (negative because the box falls this fr below the point fro which it started). Putting all that together in the equation for displacement:

Δx = v₀t +
`(1)/(2)at^2` and we are solving for time:

`-13.30=0t+(1)/(2)(-10)t^2` and

`t=\sqrt{(2(-13.30))/(-10) }` so

t = 1.6 seconds to reach the bottom of the slope from 13.30 m high.