A 10-kg crate is pulled along a horizontal floor by a constant force exerted by a person, Fp = 80N, which acts at a 45◦ angle. The floor is rough and there is a kinetic friction force of 4.1 N acting on the crate as it slides. The crate is pulled for a distance of 10 meters. a) Determine the work done by the pulling force and friction force. b) If the crate starts at rest, what is the final speed of the crate?
I have really been struggling with the second part of this problem to find the final speed. If anyone could help that would be awesome! Thank you!
Fp horizontal = 80 cos 45 = 56.6 Newtons
Fp up = 56.6 N
normal force = 10 * 9.81 - 56.6 = 41.5 N (not needed but you should know)
friction force = - 4.1 N
Work done by pull = horizontal pull force * distance in direction of pull
= 56.6 * 10 = 566 Joules
Work done by friction horizontal friction force * distance in direction of force
= - 4.1 * 10 = -41 N
final Ke of block = (1/2) m v^2 = 566 Joules - 41 Joules
so
(1/2)(10)v^2 = 525 Joules
Thank you so much! I appreciate it!
You are welcome.
To find the final speed of the crate, we can use the work-energy principle, which states that the work done on an object is equal to the change in its kinetic energy.
a) We'll begin by finding the work done by the pulling force and the friction force.
1. Work done by the pulling force (Fp):
The work done by a force is given by the formula: work = force × distance × cos(angle).
Given:
Force (Fp) = 80 N
Distance = 10 m
Angle = 45°
Using the formula, we can calculate the work done by the pulling force:
work_pull = Fp × distance × cos(angle)
= 80 N × 10 m × cos(45°)
1 Newton (N) is equivalent to 1 kg·m/s². Using this information, we can simplify the equation:
work_pull = 80 kg·m/s² × 10 m × cos(45°)
= 800 kg·m²/s² × cos(45°)
= 800 × (√2/2) kg·m²/s²
= 400 kg·m²/s²
Therefore, the work done by the pulling force is 400 kg·m²/s².
2. Work done by the friction force:
The friction force acts in the opposite direction of motion, so its work is negative:
work_friction = -friction_force × distance
Given:
Friction force = 4.1 N
Distance = 10 m
work_friction = -4.1 N × 10 m
= -41 kg·m²/s²
Therefore, the work done by the friction force is -41 kg·m²/s².
b) Now, let's calculate the final speed of the crate using the work-energy principle.
The work done on the crate is the sum of the work done by the pulling force and the work done by the friction force:
total work = work_pull + work_friction
= 400 kg·m²/s² + (-41 kg·m²/s²)
= 359 kg·m²/s²
According to the work-energy principle, this work is equal to the change in kinetic energy of the crate:
total work = ΔKE
The initial kinetic energy (KE_initial) is zero since the crate starts at rest. Therefore, we can rewrite the equation as:
total work = KE_final - KE_initial
359 kg·m²/s² = KE_final - 0
Hence, the final kinetic energy (KE_final) is 359 kg·m²/s².
The kinetic energy can be expressed as KE = 1/2 × mass × velocity². Rearranging the formula, we have:
velocity = √(2 × KE / mass)
Given:
Mass = 10 kg
KE_final = 359 kg·m²/s²
Substituting the values into the equation, we can calculate the final speed:
velocity = √(2 × 359 kg·m²/s² / 10 kg)
= √(718 kg·m²/s² / 10 kg)
= √(71.8 m²/s²)
= 8.47 m/s
Therefore, the final speed of the crate is approximately 8.47 m/s.