Final velocity of a mass moving through friction

In summary, friction does work in this distance, increasing the kinetic energy of the object by 5v^2.
  • #1
Jrlinton
134
1

Homework Statement


A block of 10 kg is pulled by a force of 100 N at and angle 30 degrees above the horizonal through a distance of 4 meters. Kinetic friction has a coefficient of 0.3.
How much work does friction do in that distance?
If its speed is 3 m/s at point a, what is its speed at point b?

Homework Equations

The Attempt at a Solution


For the first question I calculated the work of friction to be :
-u(Fn)(d)
(-0.3)((10kg*9.81m/s/s)-(100sin30))(4m)
-57.72 J

Now for the second part I took the kinetic energy of the mass moving at 3 m/s and added the work done by friction and set that equal to the final kinetic energy.
(0.5)(m)(v^2)+Wf=(0.5)(m)(v^2)
(0.5)(10kg)((3m/s)^2)-57.72J=(0.5)(10kg)(v^2)
-12.72 J= 5v^2
This is where i got stuck as the work done by friction was greater than the energy of the system to begin with. So according to this the object stopped before it even reached point b.
 
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  • #2
That is the work of friction as it is
(-u)(normal force)(d)
u=0.3
d=4.0m
Normal force= (m*g)-(FsinΘ)
with
m=10kg
g=9.81m/s/s
Magnitude of force= 100N
Angle of force above horizontal= 30 degrees
 
  • #3
Um, where are positions a and b? Without a diagram or clear description it's hard to proceed.

Presumably the 100 N force makes some contribution to the horizontal motion, not just alleviating some of the friction force. So how much energy does it contribute to the scenario?
 
  • #4
Sorry point a is x=o, point b x=4m
gneill said:
Presumably the 100 N force makes some contribution to the horizontal motion, not just alleviating some of the friction force. So how much energy does it contribute to the scenario?
So:
FcosΘ*d+(0.5)(m)(v^2)+Work of friction=(0.5)(m)(vf^2)
100cos30*4+(0.5)(10kg)(3^2)-60=(0.5)(m)(v^2)
346.41+45-60=5v^2
v=8.14 m/s
 
  • #5
Okay, the result looks good. Mind you, the lonely "kg" unit on the second line looks out of place without units being given for all the variables. So you may want to clean up the work before handing it in. Either include units for everything all the way though, use symbols for all values, or state that you're working with magnitudes and that units are ignored through the working (but definitely include the units on any final values!).
 

FAQ: Final velocity of a mass moving through friction

What is the definition of final velocity?

Final velocity is the speed and direction of an object at the end of its motion. It is the result of all the forces acting on the object, including friction.

How does friction affect the final velocity of a moving object?

Friction is a force that opposes the motion of an object. As a result, it can decrease the final velocity of a moving object by slowing it down. This is because friction converts some of the object's kinetic energy into heat.

What factors can influence the amount of friction on a moving object?

The amount of friction on a moving object can be influenced by factors such as the type of surface the object is moving on, the weight and surface area of the object, and the speed at which the object is moving.

How can the final velocity of a moving object be calculated with friction?

To calculate the final velocity of a moving object with friction, you can use the equation vf = vi + at - μkmg, where vf is the final velocity, vi is the initial velocity, a is the acceleration, μk is the coefficient of kinetic friction, m is the mass of the object, and g is the acceleration due to gravity.

Can the final velocity of a moving object ever be greater than the initial velocity?

No, the final velocity of a moving object can never be greater than the initial velocity in the presence of friction. This is because friction always acts in the opposite direction of the object's motion, causing it to slow down.

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