Final answer:
a) The net force acting on the suitcase is 40.31 N.
b) The horizontal component of the applied force is 43.30 N.
c) The work done by the applied force is 623.48J.
d) The coefficient of friction between the suitcase and the floor is 0.6662.
Step-by-step explanation:
a) The net force is determined by resolving the applied force into horizontal and vertical components. The horizontal component is F _horizontal = F⋅cos(θ) where F is the applied force and θ is the angle.
Thus, F_horizontal = 50N ⋅ cos(30∘) ≈ 43.30N
b) The horizontal component of the applied force is found by F _horizontal = F ⋅ cos(θ),
resulting in F_horizontal = 50N ⋅ cos(30∘) ≈ 43.30N.
c) The work done by the applied force is calculated using Work = Force × Distance×cos(θ), where θ is the angle between the force and the direction of motion. Substituting the given values, we get
Work = 43.30N × 15m × cos(30∘) ≈ 623.48J.
d) The coefficient of friction (μ) can be determined using =Frictional Force/Normal Force
Since the suitcase is on a horizontal surface, the normal force equals the weight of the suitcase (65N).
The frictional force is the horizontal component of the applied force (43.30N).
Therefore, μ = 43.30N/65 N ≈ 0.6662
a) The net force acting on the suitcase is 40.31 N, obtained by combining the horizontal and vertical components of the applied force.
b) The horizontal component of the applied force is 43.30 N, calculated using the cosine of the angle of pull. The work done is 623.48 J, and the coefficient of friction is approximately 0.6662, indicating a moderately rough surface.