An applied force accelerates a 50-kg crate along a frictionless floor from rest to 5 m/s.The work performed by this force is most nearly
125 J
250 J(x)
500 J
625 J
1,250 J
Can someone please explain this to me and help? I thought it was 250 J but my calculation was wrong.
W = ΔKE =KE2 –KE1 = m•v²/2 – 0 = m•v²/2 =50•25/2 =625 J
To find the work done by the applied force, we can use the work-energy principle. The work-energy principle states that the work done on an object is equal to the change in its kinetic energy.
Given that the crate starts from rest and accelerates to a final velocity of 5 m/s, we can find the change in kinetic energy using the formula:
ΔKE = (1/2) * m * (vf^2 - vi^2)
Where:
ΔKE = change in kinetic energy
m = mass of the crate (50 kg)
vf = final velocity (5 m/s)
vi = initial velocity (0 m/s, since the crate starts from rest)
Plugging in the values, we have:
ΔKE = (1/2) * 50 kg * (5 m/s)^2 - (0 m/s)^2
= (1/2) * 50 kg * (25 m^2/s^2)
= 625 J
Therefore, the work performed by the applied force is 625 J.
The closest option to this answer is 625 J, so the correct answer is 625 J.
To determine the work performed by the applied force, we can use the work-energy principle. The work done by a force is equal to the change in kinetic energy of the object it acts upon.
The formula for work is:
Work = Force * Distance * cos(theta)
In this case, the force and distance are not specified, but we can still solve the problem using the work-energy principle.
The work done by the applied force is equal to the change in kinetic energy of the crate. The initial kinetic energy is zero because the crate is at rest. The final kinetic energy is given by:
Kinetic Energy = (1/2) * Mass * Velocity^2
Plugging in the values:
Kinetic Energy = (1/2) * 50 kg * (5 m/s)^2
= (1/2) * 50 kg * 25 m^2/s^2
= 625 J
Therefore, the work done by the applied force is 625 J.
Comparing this with the answer choices provided, the closest answer is 625 J.