Question

A luggage handler pulls a 20.0 kgkg suitcase up a ramp inclined at 32.0 ∘∘ above the horizontal by a force F⃗ F→ of magnitude 169 NN that acts parallel to the ramp. The coefficient of kinetic friction between the ramp and the incline is μkμk = 0.300. The suitcase travels 3.80 mm along the ramp. calculate:

A) the work done on the suitcase by force F.
B) the work done on the suitcase by the gravitational force.
C) the work done on the suitcase by the normal force.
D) the work done on the suitcase by the friction force.
E) the total work done on the suitcase.
F) if the velocity of the suitcase is zero at the bottom of the ramp,what is its velocity after it has travelled 4.60 m along the ramp.

Answer 1

A luggage handler pulls a suitcase up a ramp inclined at 32.0 degrees above the horizontal. We need to calculate the work done on the suitcase by force F, the gravitational force, the normal force, the friction force, and the total work done on the suitcase. We also need to find the velocity of the suitcase after it has travelled a certain distance along the ramp.

Step-by-step explanation:

To calculate the work done on the suitcase, we need to consider the different forces acting on it. Let's break it down:

1. A) The work done on the suitcase by the force F is given by the equation: W = F * d * cos(θ), where d is the displacement along the ramp and θ is the angle between the force and the displacement. In this case, the force and the displacement are parallel, so the angle is 0 degrees. Therefore, the work done by force F is: W = 169 N * 3.8 m * cos(0°) = 642.2 J.
2. B) The work done by the gravitational force is given by the equation: W = m * g * d * cos(θ), where m is the mass, g is the acceleration due to gravity, d is the displacement, and θ is the angle between the force and the displacement. In this case, the gravitational force is acting vertically, so the angle between the force and the displacement is 90 degrees. Therefore, the work done by the gravitational force is: W = 20 kg * 9.8 m/s2 * 3.8 m * cos(90°) = 0 J.
3. C) The work done by the normal force is zero since the normal force is perpendicular to the displacement.
4. D) The work done by the friction force is given by the equation: W = - μk * Fn * d * cos(θ), where μk is the coefficient of kinetic friction, Fn is the normal force, d is the displacement, and θ is the angle between the force and the displacement. In this case, the angle between the friction force and the displacement is 180 degrees, and the normal force is equal to the gravitational force, which is 20 kg * 9.8 m/s2. Therefore, the work done by the friction force is: W = - 0.3 * (20 kg * 9.8 m/s2) * 3.8 m * cos(180°) = -225.36 J.
5. E) The total work done on the suitcase is the sum of the work done by each force: Wtotal = WF + Wg + WN + Wfr. In this case, WN and Wg are both zero, so Wtotal = WF + Wfr = 642.2 J + (-225.36 J) = 416.84 J.
6. F) To find the velocity of the suitcase after travelling 4.60 m along the ramp, we can use the work-energy principle: the work done on an object is equal to the change in its kinetic energy. The initial kinetic energy of the suitcase is zero since its velocity is zero. So the work done on the suitcase is equal to the final kinetic energy: Wtotal = (1/2) * m * vf2. Rearranging the equation, we can solve for the final velocity vf: vf = sqrt((2 * Wtotal) / m) = sqrt((2 * 416.84 J) / 20 kg) = 2.52 m/s.

Answer 2

a) W₁ = 8242.2 J, b) W₁ = 8242.2 J , c) W₃ = 0 , d) W₄ = -189.51 J ,

f) v = 27.24 m / s

Step-by-step explanation:

a) Work is defined by

W = F. d ​​= F d sin θ

where angle is between force and displacement

n this case the suitcase is going up and the outside F is parallel to the plane, so the angle is zero and the cosine is 1

W = F d

Let's calculate

W = 169 3.8

W₁ = 8242.2 J

b) the gravitational force is vertical so it has an angle with respect to the horizontal parallel to the plane of

θ’= 90 - θ

θ'= 90-32 = 58º

W = m g d thing θ ’

W = 20 9.8 3.8 thing (180 + tea ’) =

W = 744.8 cos (180 + 32)

W₂ = -631.6 J

c) The normal work, as it has 90º with respect to the displacement, its work is zero

W₃ = 0

d) the work of the friction force

Let's write Newton's second law the Y axis

N- Wy = 0

Cos 32 = Wy / W

N = W cos 32

The expression for friction force is

fr = μ N

fr = μ mg cos 32

fr = 0.300 20 9.8 cos (32)

fr = 49.87 N

The work of the friction force

W = fr d cos 180

W₄ = -49.87 3.8

W₄ = -189.51 J

E) The total work

W = W₁ + W₂ + W₃ + W₄

W = 8242.2- 631.6 + 0 -189.51

W_total = 7421.09 J

F) Usmeosel theorem of work and energy

W = ΔK

W = ΔK = ½ m v² - 0

v =√ 2W / m

v = √ (2 7421.09 / 20)

v = 27.24 m / s