A worker pushed a 18.0 kg block 12.0 m along a level floor at constant speed with a force directed 28.0° below the horizontal. If the coefficient of kinetic friction between block and floor was 0.450, what were (a) the work done by the worker's force and (b) the increase in thermal energy of the block-floor system?

(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:

- F cos θ = 0, = μk N

where

= the force friction (n)

μk = he coefficient of kinetic friction

F = force (N)

N = normal force (N)

θ = angle

thus

= 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 = d

where

Δ = the change of thermal energy (J)

= F cos θ

   = 118.18 cos 28°

   = 104.35 N

so,

Δ = d

        = (104.35) (12)

        = 1252.2 J

answer is c

explanation: