A 10kilogram box is pushed along a rough horizontal surface...

A 10kilogram box is pushed along a rough horizontal surface by a constant horizontal force F as shown above. At time t = 0s, the velocity of the box is 6.0 m/s in the same direction as the force. At time t = 12s the velocity of the box is 30 m/s. The coefficient of sliding friction is 0.2. a. What is the magnitude of the force F? b. At t = 12s, two 3-kg books are dropped into the box. If the applied force stays constant, what is the speed of the box at time t = 20s?

Transcript

00:01 In this problem, we have a block that is moving on a horizontal surface that has friction.

00:08 And so the mass of this box is 10 kilograms.

00:13 We're told the coefficient of friction on this rough surface is 0 .2.

00:17 It says the box is initially moving at 6 meters per second.

00:23 So that's at time 0.

00:24 And then 12 seconds later, it's going to be moving at 30 meters per second.

00:29 So the final velocity for this 12 second interval is 30 meters per second.

00:38 So the first thing we need to do, we're trying to actually find out what is the force applied during this time.

00:45 So we're going to need to do a free body diagram.

00:47 So this is for a.

00:49 Let me do a free body diagram first.

00:52 We know that pointing down is always force of gravity.

00:56 And pointing up is normal force.

00:59 Now normal force is not always up.

01:01 It is perpendicular to the surface.

01:02 And since our surface is horizontal, it's going to be pointing directly up.

01:06 We have a force applied, and then we're going to have a force of friction.

01:13 Now we know the force of friction is not as big as the force applied, because from the diagram you can see that it is moving in the same direction.

01:20 So let's look at this.

01:21 It is only moving in the horizontal direction.

01:24 It is not moving up, nor is it moving down.

01:26 So that means the net force in the y direction is equal to zero.

01:34 Now we know for the x direction, it is moving and it is accelerating, so it is mass times acceleration.

01:42 So these are the two equations we're going to need to use.

01:45 So we're trying to find the force applied, which is right here, and that is in our x direction.

01:51 So this is the equation we're going to need to use.

01:54 F net equals mass times acceleration.

01:58 So let's put in all the forces in the x direction.

02:01 We have that force applied, which is what we're looking for, and we have a force.

02:05 Friction in the opposite direction, which is why i have it as negative.

02:09 And then we have mass and acceleration.

02:12 So there's several steps we're going to have to do to get everything we need.

02:15 The first step.

02:17 I'm going to need to find acceleration.

02:20 And so i'm going to use a constant acceleration, which is final velocity equals initial velocity plus acceleration times time.

02:29 So we have 30 equals six plus acceleration times 12.

02:35 Subtract six from both.

02:36 Sides and you get 24 equals acceleration times 12 divide both sides by 12 to get acceleration by itself and you get an acceleration of two meters per second squared so we can plug that in here the only problem is we do not have force of friction but because we were given this we can solve for our force of friction with the coefficient of friction equation so let's scroll down and get a little bit more room so for our force of friction equation, let's go ahead and block that off.

03:14 It is force of friction over normal force is equal to the coefficient of friction.

03:20 Multiply both sides by normal force to get it by itself.

03:25 And that gives us a force of friction is equal to the coefficient of friction times normal force.

03:31 The problem is we don't have normal force.

03:34 But we already saw up here that some of the forces in the y direction are equal to zero.

03:41 So that means normal force minus force of gravity equals zero...

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