Solution of Problem 42, Chapter 8, Fundamentals of Physics 9th - Halliday & Resnick

Solution of Problem 42, Chapter 8

Fundamentals of Physics 9th - Halliday & Resnick

Question:

A worker pushed a 27 kg block 9.2 m along a level floor at constant speed with a force direction 32° below the horizontal. If the coefficient of kinetic friction between block and floor was 0.20, what were (a) the work done by the worker's force and (b) the increase in thermal energy of the block - floor system?

Answer:

a) Work done by worker's force:

    Work =  force * distance * cos 𝚹

    Force = 𝚺F = ma = m (g - friction force)

    friction force = μk * N ( Where N is the normal force)

    N = mg

    Substituting the values

    Work = (27 kg) (9.8 m/s^2 - 0.2 * 9.8 m/s^2) * 9.2 m * cos(32°)

    Work = 227.6 J

b) Increase in thermal energy:

    ΔQ = work done by friction force =  friction force * distance

    friction force = μk * N = μk * m * g

    Substituting the values:

    ΔQ = 0.2 * 27 kg * 9.8 m/s^2 * 9.2 m

    ΔQ = 486.4 J