Answer:
(a) The work done by the worker's force is 1252.2 J.
(b) The increase in thermal energy of the block-floor system is 1252.2 J.
Explanation:
given information
mass of the block, m = 18 kg
the distance, d = 12 m
angles, θ = 28°
the coefficient of kinetic friction, μk = 0.45
(a) the work done by the worker's force
according the newton's first law
horizontal force:
[tex]f_{k}[/tex] - F cos θ = 0, [tex]f_{k}[/tex] = μk N
where
[tex]f_{k}[/tex] = the force friction (n)
μk = he coefficient of kinetic friction
F = force (N)
N = normal force (N)
θ = angle
thus
[tex]f_{k}[/tex] = F cos θ
μk N = F cos θ
N = F cos θ/μk
then for the vertical force:
N - mg - F sin θ = 0
where
m = mass (kg)
g = gravitational force (9,8 m/s²)
thus,
N = mg + F sin θ
F cos θ/μk = mg + F sin θ
F cos θ/μk - F sin θ = m g
F (cos θ/μk - sin θ) = m g
F = m g/(cos θ/μk - sin θ)
= (18 x 9.8)/ (cos 28°/0.450 - sin 28°)
= 118.18 N
next, we can calculate the work done by the worker's force by the following formula
W = F d cos θ
where
W = work (J)
d = distance (m)
W = (118.18) (12) cos 28°
= 1252.2 J
b) the increase in thermal energy of the block-floor system?
Δ[tex]E_{th}[/tex] = [tex]f_{k}[/tex] d
where
Δ[tex]E_{th}[/tex] = the change of thermal energy (J)
[tex]f_{k}[/tex] = F cos θ
= 118.18 cos 28°
= 104.35 N
so,
Δ[tex]E_{th}[/tex] = [tex]f_{k}[/tex] d
= (104.35) (12)
= 1252.2 J
