Full transcript
0:00good morning a 2.0 kg Mass slides to the
0:04right 80.0 CM along a surface and comes
0:08to rest in 0.50 seconds what net force
0:11is acting on the mass as it slides along
0:14the
0:18surface so we know the mass of the
0:20object is 2.0 kg we know the
0:22displacement is 80.0 CM the change in
0:25time is 0.50 seconds and we are solving
0:28for the net force first thing I would
0:30suggest that you do is convert the
0:32displacement from centimeters to meters
0:35it's always a good idea to start by
0:37putting everything into base SI
0:41Dimensions 100 cm equal 1 M so 80 cm is
0:45equal to 0.80
0:47M next we should draw the free body
0:51diagram because it's a problem that
0:53deals with forces so you need a free
0:55body diagram
0:59[Music]
1:01the force normal is always normal to the
1:03surface and a push so it is up
1:05perpendicular to the surface the force
1:07of gravity is always straight down and
1:10we know there is a force of friction
1:12because it is sliding to a stop
1:14therefore it is accelerating in the X
1:15Direction therefore there must be some
1:18Force causing it to slow down which is
1:20the force of friction because it's
1:22sliding to the right we know the force
1:24of friction is opposing that motion so
1:25the force of friction is going to be to
1:28the left now now that we've drawn our
1:30free body diagram we should use Newton's
1:32Second Law and we should sum the forces
1:34in let's start with the y
1:40direction the net force in the y
1:43direction equals the force normal which
1:44is positive because it's up minus the
1:47force of gravity which is negative
1:48because it's down and the net force is
1:50always equal to mass times the
1:52acceleration in this particular case
1:54this is the acceleration in the y
1:55direction because we have sum the forces
1:57in the y direction we know the mass is
2:00sliding along a horizontal surface to
2:02the right therefore the acceleration in
2:04the y direction is equal to zero it's
2:06not moving up or down at all therefore
2:09the net force is equal to zero which
2:12means that the force normal minus the
2:14force of gravity is equal to zero which
2:16means the force normal if we add the
2:17force of gravity to both sides the force
2:19normal is equal to the force of gravity
2:21and the equation for the force of
2:22gravity is mass times the acceleration
2:24due to gravity now I'm actually not
2:26going to plug anything in for that right
2:28now because you'll discover that this
2:30was completely useless as far as this
2:32problem is concerned um it's useful to
2:35do in the long run cuz we're going to do
2:37this a lot more as we get more into this
2:40class but uh for this particular problem
2:42that actually doesn't do us any good
2:47sorry next we can sum the forces in the
2:50X Direction the net force in the X
2:52direction if you look the only force
2:54acting in the X direction is the force
2:55of friction and it's to the left
2:57therefore it's negative therefore the
2:59net force for is equal to negative force
3:01of friction which is equal to because
3:02Newton's Second Law the net force is
3:04always equal to mass time the
3:05acceleration in this particular case
3:07because we're summing the forces in the
3:08X Direction This is equal to masstimes
3:10the acceleration in the X
3:13Direction our goal is to solve for the
3:15net force we have the mass of the object
3:19it's 2 kg we simply need the
3:21acceleration of the object now we know
3:24that this is a constant force the force
3:26of friction is going to be constant
3:28therefore the acceleration is going to
3:30be constant the acceleration is going to
3:32be uniform therefore we can use the
3:34uniformly accelerated motion equations
3:36to solve for the
3:39acceleration one important piece of
3:41information from the problem is that it
3:44comes to a rest in 0.50 seconds
3:47therefore we also know that the final
3:50velocity is equal to
3:52zero therefore we can use the uniformly
3:55accelerated motion equation the
3:56displacement is equal to 1/2 * the
3:58quantity velocity velocity initial plus
4:00velocity final times the change in time
4:03we can substitute in our known
4:06values the displacement is 0.8 we have
4:091/2 * the quantity velocity initial we
4:11do not know plus the velocity final
4:13which is zero times the change in time
4:15which is 0.5 seconds so now we can solve
4:17for the initial
4:21velocity multiplying 12 by 0.5 gives us
4:240.25 * the velocity initial dividing
4:27both sides by 0.25 gives us the velocity
4:30initial equals 0.8 / 0.25 or 3.2 m/s now
4:36we can solve for the acceleration using
4:38a different UAM
4:45equation the velocity final equals the
4:47velocity initial plus the acceleration
4:49times the change in time velocity final
4:51is zero velocity initial is 3.2 and the
4:54change in time is 0.5 uh subtracting 3.2
4:57from both sides gives us 0 - 3 .2 is
5:00equal to 0.5 * the acceleration
5:02therefore dividing by 0.5 gives us the
5:04acceleration equal 0 - 3.2 / 0.5 or -6.4
5:10m/s squared that is the acceleration in
5:13the X
5:14Direction now we can substitute in mass
5:17time acceleration to get the net force
5:19in the X
5:21Direction mass is 2 the acceleration in
5:25the X direction is -6.4 m/s squ so
5:28that's equal to 12.8 or with two
5:32significant digits -3
5:35Newtons thank you very much for learning
5:37with me today I enjoyed learning with
5:39you