3813.75 Joules of work is being done on a 150 lb object while it is moved 75 ft across a horizontal plane, what is the final velocity if the coefficient of kinetic friction is 0.25 and the initial velocity is 1 ft / s?
vf^2=vi^2+2ad
but a=force/mass=mu*mg/m=mu*g
solve for vf
change lb to kg, 75 f to meters. vi to m/s
Thanks for the help, but I still need help on more one piece.
How do I get the final velocity or get the ^2 out of the equation. Here are my following numbers:
Vi= 0.3048 m/s
Vf= ?
d = 22.86 m
a = 166 m/s^2
t = Not required
Based on the numbers I have would I set it up this way???
Vf^2= 0.3048^2 + 2*(166*22.86)
This process comes out to a impossible number... Please help!
To find the final velocity of the object, we need to consider the work done, the displacement, and the coefficient of kinetic friction.
First, let's convert the mass of the object from pounds to kilograms. We know that 1 lb is approximately equal to 0.4536 kg, so the mass of the object is:
mass = 150 lb × 0.4536 kg/lb = 68.04 kg
Next, we can use the work-energy principle to determine the final kinetic energy of the object.
The work done on the object is equal to the change in its kinetic energy. The work-energy principle can be expressed as:
work = change in kinetic energy
In this case, the work done is given as 3813.75 Joules, so we can write the equation as:
3813.75 J = change in kinetic energy
Since the initial velocity is given as 1 ft/s, we can express it in meters per second:
initial velocity = 1 ft/s × 0.3048 m/ft = 0.3048 m/s
The final kinetic energy can be expressed as:
final kinetic energy = 0.5 × mass × final velocity^2
Let's assume the final velocity is v (in m/s) and set up the equation for the work-energy principle:
3813.75 J = 0.5 × 68.04 kg × (v^2 - 0.3048 m/s)^2
Now, we can use the equation for frictional force to find the force of kinetic friction. The equation is given by:
frictional force = coefficient of kinetic friction × normal force
The normal force can be calculated by multiplying the mass of the object by the acceleration due to gravity (9.8 m/s²):
normal force = mass × gravity = 68.04 kg × 9.8 m/s²
Substituting the given coefficient of kinetic friction of 0.25:
frictional force = 0.25 × (68.04 kg × 9.8 m/s²)
The work done against friction is equal to the frictional force multiplied by the distance traveled:
work against friction = frictional force × displacement = 0.25 × (68.04 kg × 9.8 m/s²) × 75 ft × 0.3048 m/ft
Now, we can subtract the work done against friction from the total work done to find the network done on the object:
work against friction = (0.25 × 68.04 kg × 9.8 m/s² × 75 ft × 0.3048 m/ft)
network = 3813.75 J - (0.25 × 68.04 kg × 9.8 m/s² × 75 ft × 0.3048 m/ft)
Finally, we can solve for the final velocity by equating the network to the change in kinetic energy and solving for v:
network = 0.5 × 68.04 kg × (v^2 - 0.3048 m/s)^2
By rearranging the equation and solving for v, we can find the final velocity of the object.