a delivery person pulls a 20.8 kg box across the floor. the force exerted on the box is 95.6n (35 above the horizontal). the force of kinetic friction on the box has a magnitude of 75.5n. the box starts from rest. using the work-energy theorem determine the speed of the box after being dragged 0.750m.
work done by person ... 95.6 N * cos(35º) * .750 m
work done by friction ... 75.5 N * .750 m
the difference in the two work amounts is the kinetic energy of the box
K.E. = 1/2 * mass * speed^2
Well, looks like we need to do some work to find the speed of this box! And by "work," I don't mean the funny kind... I mean actual physics work.
According to the work-energy theorem, the net work done on an object is equal to the change in its kinetic energy. In this case, the net work done on the box would be the work done by the applied force minus the work done by friction.
To calculate the work done by the applied force, we multiply the magnitude of the applied force (95.6 N) by the displacement of the box (0.750 m). However, since the applied force is at an angle of 35 degrees above the horizontal, we need to consider only the horizontal component of the force. So, the work done by the applied force would be:
Work_applied = 95.6 N * cos(35 degrees) * 0.750 m
To calculate the work done by friction, we multiply the magnitude of the frictional force (75.5 N) by the displacement of the box (0.750 m), because the force of friction and displacement are in the same direction. So, the work done by friction would be:
Work_friction = 75.5 N * 0.750 m
Now, since the net work done is equal to the change in kinetic energy, we have:
Net Work = Change in KE
Work_applied - Work_friction = (1/2) * mass * (final velocity^2 - initial velocity^2)
The box starts from rest, so the initial velocity is zero. The mass of the box is 20.8 kg.
Now let's plug in the values, crunch some numbers, and see what pops out as the final velocity of the box!
To determine the speed of the box after being dragged 0.750m using the work-energy theorem, we need to calculate the net work done on the box.
The work done on an object can be calculated using the equation:
Work = Force x Distance x cos(theta)
Where:
- Force is the applied force on the box (95.6N)
- Distance is the distance the box is dragged (0.750m)
- Theta is the angle between the force and the direction of motion (35 degrees above the horizontal)
First, let's calculate the work done by the applied force:
Work_applied = 95.6N x 0.750m x cos(35°)
Next, let's calculate the work done by the kinetic friction force:
Work_friction = 75.5N x 0.750m x cos(180°)
Note that the angle between the friction force and the direction of motion is 180 degrees because the friction force acts in the opposite direction of motion.
The net work done can be calculated by subtracting the work done by the friction force from the work done by the applied force:
Net work = Work_applied - Work_friction
Finally, we can use the work-energy theorem to determine the speed of the box:
Net work = Change in kinetic energy
Change in kinetic energy = 1/2 x mass x (final velocity^2 - initial velocity^2)
Since the box starts from rest, the initial velocity is zero.
Net work = 1/2 x mass x (final velocity^2 - 0)
By equating the net work to the change in kinetic energy, we can solve for the final velocity of the box.
Net work = 1/2 x mass x final velocity^2
Substituting the calculated values into the equation, we can determine the final velocity of the box.
To determine the speed of the box using the work-energy theorem, we need to calculate the net work done on the box and equate it to the change in its kinetic energy. The net work is the sum of the work done by the applied force and the work done by friction.
First, let's find the work done by the applied force. The work done by a force can be calculated using the formula:
Work = Force × Distance × cos(θ)
Where:
- Work is the work done by the force
- Force is the magnitude of the force exerted on the box (95.6 N)
- Distance is the distance the box is dragged (0.750 m)
- θ is the angle between the force and the direction of motion (35 degrees above the horizontal)
Note that we need to calculate the horizontal component of the force to find the work done. Since the force makes an angle of 35 degrees above the horizontal, the horizontal component can be found using the equation:
Force_horizontal = Force × cos(θ)
Now, let's calculate the work done by the applied force:
Work_applied = Force_horizontal × Distance
= (Force × cos(θ)) × Distance
= (95.6 N × cos(35°)) × 0.750 m
Next, let's find the work done by friction. The work done by friction can be calculated using the formula:
Work_friction = Force_friction × Distance
Given that the magnitude of the force of kinetic friction is 75.5 N, we can calculate the work done by friction:
Work_friction = 75.5 N × 0.750 m
Now, let's calculate the net work done:
Net_work = Work_applied - Work_friction
Finally, we can equate the net work done to the change in kinetic energy using the work-energy theorem:
Net_work = ΔKE
The change in kinetic energy, ΔKE, is equal to the final kinetic energy of the box because it starts from rest. Thus, we have:
ΔKE = (1/2)mv^2
Where:
- m is the mass of the box (20.8 kg)
- v is the speed of the box
Now, we can solve the equation to find the speed of the box after being dragged 0.750 m.