Smorse The crate starts from rest and is towed by the force...

smorse The crate starts from rest and is towed by the force $F = 150 \text{ lbf}$. If the coefficient of kinetic friction between the 100 lbf crate and the ground is $\mu_k = 0.25$, 8. Determine the speed of the crate when $t = 4s$ A 98.0 ft/s B 57.1 ft/s C 68.3 ft/s D 159 ft/s 9. Determine the distance the crate travels when $t = 4s$ A 114 ft B 137 ft C 318 ft D 195 ft (Work Energy, Linear Impulse momentum) F 30°

Transcript

00:01 Hi all, here in this case, the free body diagram is as follows.

00:06 So this is the free body diagram.

00:09 Here, there is normal reaction and velocity v9 meter per second.

00:14 There is forced due to friction.

00:16 This is weight mg.

00:17 It is resolved to mg cos theta and mg sine theta given the coefficient of static friction is equal to 0 .4 and the coefficient of kinetic friction is equal to 0 .25.

00:31 By equation, we can see that here the net force is equal to here we have got mg sine theta acting right side plus the force due to friction which is also acting right side.

00:45 Or we can say that net force can be represented as mass times acceleration a is equal to mg sine theta plus instead of frictional force we can write mu k into mg into cos theta.

01:00 So this is the equation to be used here.

01:02 Here m is common in all the terms so canceling m it becomes a equal to g sine theta plus mu k into cos theta or we can say that here this is equal to 9 .8 meter per second square into sine 20 degree plus here there is also a term g here mu k into g into cos theta value of mu k here is 0 .25 into g is 9 .8 meter per second square into course 20 degree.

01:42 Upon solving this we will get acceleration a is equal to 5 .654 meter per second square and here the direction is towards the downward direction.

01:57 Next let us find the time taken for the crate to stop.

02:02 Here the time taken for the crate to stop is given by the equation time t is equal to velocity divided by acceleration a here the velocity is given which is 9 meter per second divided by acceleration a is 5 .654 meter per second square upon solving this we will get time t is equal to 1 .59 seconds here in this case when the crate stop what happens is the gravitational force acts on the crate in downward direction.

02:46 Thus, we can say that here the force must be less than static friction or static friction force here...

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