A block with a mass of m = 3.12 kg is at rest on an incline....

A block with a mass of m = 3.12 kg is at rest on an incline. The angle of the incline is ? = 26.5° with respect to the horizontal. The coefficient of static friction between the object and the incline is ?s = 0.763, the coefficient of kinetic friction is ?k = 0.564. m ? What would be the magnitude of the object's acceleration after the object is set in motion downhill on the incline? (in m/s^2) A: 2.62×10^-1 B: 3.06×10^-1 C: 3.59×10^-1 D: 4.20×10^-1 E: 4.91×10^-1 F: 5.74×10^-1 G: 6.72×10^-1 H: 7.86×10^-1 What would be the magnitude of the object's acceleration after the object is set in motion uphill on the incline? (in m/s^2) A: 5.06 B: 5.72 C: 6.47 D: 7.31 E: 8.26 F: 9.33 G: 1.05×10^1 H: 1.19×10^1

Transcript

00:01 Hello everyone, we are going to understand this question here given in the question given mass of block, mass of block that is represented by small m, which is 3 .12 kg, which is at rest on the inclined trend.

00:34 And angle of inclination that is given 26 .5.

00:41 Angle of inclination theta is equal to 26 .5 degree with respect to the horizontal.

01:01 And the coefficient of static friction that is represented by muus, which is 0 .763.

01:14 And coefficient of kinetic friction that is represented by mu k which is 0 .564.

01:25 Now what would be the magnitude of object's acceleration after the object is set to set in motion in downward direction? so when this object is released, so first of all let's draw the fvd weight of the block that is working vertically downward which is mg.

01:51 It is making theta degree.

01:54 So component of weight along to the inclined plane that is mg cos theta sorry mg sine theta if this is making theta angle and this is mg cos theta which will balance by the normal reaction so friction force will be in upward direction that is represented by f s so if mg sine theta is greater than fs, then it will accelerate else it will not.

02:47 So let acceleration be a.

02:57 So m .a is equal to mg sine theta minus fs.

03:06 And fs is equal to mu into mg cost theta.

03:13 Now, substituting the value in this here, value of mu, mu k, first of all, we will calculate the static friction.

03:25 So mub s is equal to 0 .7 .63 into value of m is 3 .1 2 into value of g is 9 .8 into cost.

03:40 Here angle is given 26 .5.

03:45 Now doing further calculation, after doing further calculation, we will get 20 .87 newton...

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