Transcript Pulling a block

A 2.60 kg mass is being pulled by a force of 19.6 N at an angle of elevation of
35.0° as shown in the diagram below. The coefficient of friction between the
floor and the block is 0.270. If the block starts from rest, what is its speed after
being pulled with this force for 11.0 s? Hint: find the net acceleration first. (23
marks)
m = 2.60 kg
q app = 35.0°
m = 0.270
q incl = 0°
g = 9.81 m
s2
Fapp = 19.6 N
t = 11.0 s
vi = 0 m
s
vf = ?
In order to solve the acceleration question, one must follow these steps:
1. Draw a free-body diagram.
2. Find the x- and y-components of the applied force vector.
3. Find the net force in the y-direction by adding the applied force in the ydirection to the weight if it’s being pushed and subtract the applied force
in the y-direction if it’s being pulled.
4. Use the net force in the y-direction to calculate the force of friction.
5. Calculate the net force in the x-direction by subtracting the frictional
force from the applied force in the x-direction.
6. Calculate the net acceleration by dividing the net force in the x-direction
by the mass of the block.
7. Calculate final velocity using kinematics equations.
To understand what is happening in this problem, one must first construct a
free-body diagram to describe the forces involved.
Fapp is the applied force acting on the mass.
Fappx is the component of the applied force acting in the
Fapp
Fappy
Ff
Fappx
Fg
x-direction.
Fappy is the component of the applied force acting in the
y-direction.
Fg is the force of gravity (weight) acting on the
mass.
Ff is the force of friction acting on the mass.
Fappy
Fapp
= sinq
In order to solve for the final speed of the block, we
must first break down the applied force into its xF
and y- components.
app
Fappy
Fappy = Fapp sinq
Fappy =19.6 N sin35.0°
Fappy =11.2 N
Because this block is being pulled, the force acting
in the y-direction is an upward force as seen here.
If the block were being pushed, the force in the ydirection would be a downward force and it would
add to the weight of the object being pushed.
Fappy
Fapp
Fappx
Fapp
= cosq
There is a component of the applied force that acts in
the direction of motion, or the x-direction. We
calculate this force by using the cos function.
Fapp
Fappx = Fapp cosq
Fappx =19.6 N cos35.0°
Fappy =16.1 N
Fappx
This component in the x-direction will be opposed
by the force of friction. However, what must first
be found is the net force acting in the y-direction.
It is necessary for us to calculate the
effective weight (net force acting in the
y-direction) of the block, given the fact
that it has its own weight MINUS the
component of the pulling force that we
have applied.
Fapp
Fappy
Ff
Fappx
Fg
The effective weight is increased when
we are pushing because we are adding
to the weight, but it is decreased when
we are pulling the mass because we out
y-component applies a lifting force to
the mass.
Fnety = Fg + Fappy
It is necessary for us to calculate the
effective weight (net force acting in the
y-direction) of the block, given the fact
that it has its own weight MINUS the
component of the pulling force that we
have applied.
Fnety = Fg - Fappy
Fnety = m g - Fappy
Fnety = ( 2.60 kg) ( 9.81m / s 2 ) -11.2 N
Fnety = 25.5 N -11.2 N
Fnety =14.3N
Fappy
Fg
Fnety
The effective weight is increased when
we are pushing because we are adding
to the weight, but it is decreased when
we are pulling the mass because we out
y-component applies a lifting force to
the mass.
Once you have calculated the net force in the
y-direction, you can then calculate the net
force acting in the x-direction because the net
force in the y-direction is needed to calculate
the force of friction.
One must finally
calculate the net
acceleration of the block
using Newton’s 2nd
Law.
Fapp
Fappy
Ff
Fappx
F netx = F appx + F f
Fnetx = Fappx - Ff
m anetx = Fappx - m m g cosq incl
Once you have calculated the net force in the
y-direction, you can then calculate the net
force acting in the x-direction because the net
force in the y-direction is needed to calculate
the force of friction.
Fappx - m m gcosqincl
anetx =
m
16.1N - ( 0.270) ( 2.60kg) 9.81 m 2
s
anetx =
( 2.60kg)
(
anetx = 3.54 m
One must finally
calculate the net
acceleration of the block
using Newton’s 2nd
Law.
) cos0°
s2
Fappx
Ff
Fnetx
Dv
anetx =
Dt
v f = vi + anetx Dt
(
At this point the question merely
becomes a kinematic question.
We solve for the final velocity beginning
with the definition of acceleration.
)
v f = 0 m + 3.54 m 2 (11.0s)
s
s
v f = 38.9 m
s
Fapp