Answer
Work = Force × Distance × cosθ
where θ is the angle between the applied force and the direction of motion, and cosθ is the cosine of that angle.
For pushing, the force required to overcome friction is:
F = μkN = μkmg = 0.20 × 50 kg × 9.81 m/s^2 = 98.1 N
where μk is the coefficient of kinetic friction, N is the normal force (equal to the weight of the crate), and m is the mass of the crate.
The force required to move the crate at a constant speed is equal and opposite to the force of friction, so the total force required to push the crate is:
Ftotal = F + ma = 98.1 N + 0 = 98.1 N
where a is the acceleration of the crate (which is zero since it is moving at a constant speed).
Using the equation for work, we can calculate the minimum work required to push the crate:
Work = Ftotal × d × cosθ = 98.1 N × 15 m × 1 = 1472.5 J
where d is the distance the crate is moved, and cosθ is 1 since the force is applied parallel to the direction of motion.
For pulling, the force required to overcome friction is the same, but the effective force applied to the crate is reduced by the angle θ. We can resolve the force into horizontal and vertical components:
Fhoriz = F × cosθ = 98.1 N × cosθ
Fvert = F × sinθ = 98.1 N × sinθ
The horizontal force is what actually moves the crate, so we can use it to calculate the acceleration:
Fhoriz = ma
a = Fhoriz / m = (98.1 N × cosθ) / 50 kg
The force required to move the crate at a constant speed is again equal and opposite to the force of friction, so the total force required to pull the crate is:
Ftotal = Fhoriz + ma = 98.1 N × cosθ + [(98.1 N × cosθ) / 50 kg] × 15 m × sinθ
Using the equation for work, we can calculate the minimum work required to pull the crate:
Work = Ftotal × d × cosθ = [98.1 N × cosθ + (98.1 N × cosθ) / 50 kg × 15 m × sinθ] × 15 m × cosθ
For the specific angle θ given in the problem, we can substitute in the values and calculate the work required:
Work = [98.1 N × cos(30°) + (98.1 N × cos(30°)) / 50 kg × 15 m × sin(30°)] × 15 m × cos(30°)
Work ≈ 1452 J
Therefore, the minimum work required to pull the crate is slightly less than the work required to push it.
Work Step by Step
Work = Force × Distance × cosθ
where θ is the angle between the applied force and the direction of motion, and cosθ is the cosine of that angle.
For pushing, the force required to overcome friction is:
F = μkN = μkmg = 0.20 × 50 kg × 9.81 m/s^2 = 98.1 N
where μk is the coefficient of kinetic friction, N is the normal force (equal to the weight of the crate), and m is the mass of the crate.
The force required to move the crate at a constant speed is equal and opposite to the force of friction, so the total force required to push the crate is:
Ftotal = F + ma = 98.1 N + 0 = 98.1 N
where a is the acceleration of the crate (which is zero since it is moving at a constant speed).
Using the equation for work, we can calculate the minimum work required to push the crate:
Work = Ftotal × d × cosθ = 98.1 N × 15 m × 1 = 1472.5 J
where d is the distance the crate is moved, and cosθ is 1 since the force is applied parallel to the direction of motion.
For pulling, the force required to overcome friction is the same, but the effective force applied to the crate is reduced by the angle θ. We can resolve the force into horizontal and vertical components:
Fhoriz = F × cosθ = 98.1 N × cosθ
Fvert = F × sinθ = 98.1 N × sinθ
The horizontal force is what actually moves the crate, so we can use it to calculate the acceleration:
Fhoriz = ma
a = Fhoriz / m = (98.1 N × cosθ) / 50 kg
The force required to move the crate at a constant speed is again equal and opposite to the force of friction, so the total force required to pull the crate is:
Ftotal = Fhoriz + ma = 98.1 N × cosθ + [(98.1 N × cosθ) / 50 kg] × 15 m × sinθ
Using the equation for work, we can calculate the minimum work required to pull the crate:
Work = Ftotal × d × cosθ = [98.1 N × cosθ + (98.1 N × cosθ) / 50 kg × 15 m × sinθ] × 15 m × cosθ
For the specific angle θ given in the problem, we can substitute in the values and calculate the work required:
Work = [98.1 N × cos(30°) + (98.1 N × cos(30°)) / 50 kg × 15 m × sin(30°)] × 15 m × cos(30°)
Work ≈ 1452 J
Therefore, the minimum work required to pull the crate is slightly less than the work required to push it.
