Question

13 fill in the blank text
a person pushes a 25kg bookshelf by applying +38.75n of force. they are able to get it moving with an acceleration of +0.75m/s/s across a carpet to the other side of the room.
complete the following:
(1) list the known values. (list the variable with the correct direction, magnitude, and unit of measure for each known value in the order in which they are found in the question. do not use spaces at all.)
(2) draw a force diagram on your paper.
(3) calculate the net force (f_net) acting on the bookshelf. (give the direction, magnitude, and correct unit of measure. do not use spaces at all.)
(4) calculate the bookshelfs weight (f_g). (give the direction, magnitude, and correct unit of measure. do not use spaces at all.)
(5) calculate the force of kinetic friction (f_fk). (give the direction, magnitude, and correct unit of measure. do not use spaces at all.)
(6) calculate the coefficient of kinetic friction (μk).

Explanation:

Step1: List known values

Mass \(m = 25kg\), applied force \(F_{applied}=+38.75N\), acceleration \(a = +0.75m/s^{2}\)

Step2: Calculate net force (\(F_{NET}\))

Using Newton's second law \(F_{NET}=ma\).
Substitute \(m = 25kg\) and \(a=0.75m/s^{2}\)
\(F_{NET}=25\times0.75 = 18.75N\) (direction same as acceleration, +)

Step3: Calculate weight (\(F_{g}\))

Using \(F_{g}=mg\), \(g = 9.8m/s^{2}\)
\(F_{g}=25\times9.8=245N\) (direction - as weight acts downwards)

Step4: Calculate force of kinetic friction (\(F_{FK}\))

Using \(F_{NET}=F_{applied}-F_{FK}\)
\(F_{FK}=F_{applied}-F_{NET}\)
Substitute \(F_{applied}=38.75N\) and \(F_{NET}=18.75N\)
\(F_{FK}=38.75 - 18.75=20N\) (direction - as friction opposes motion)

Step5: Calculate coefficient of kinetic friction (\(\mu_{K}\))

Assume normal force \(F_{N}=F_{g}\) (vertical equilibrium)
Using \(F_{FK}=\mu_{K}F_{N}\)
\(\mu_{K}=\frac{F_{FK}}{F_{N}}=\frac{20}{245}\approx0.0816\)

Answer:

1. \(m = 25kg,F_{applied}=+38.75N,a=+0.75m/s^{2}\)
2. (Draw force diagram with applied force +, friction -, weight -, normal +)
3. \(F_{NET}=+18.75N\)
4. \(F_{g}=-245N\)
5. \(F_{FK}=-20N\)
6. \(\mu_{K}\approx0.0816\)