Question

multiple choice question
a 54.2 - kg crate is pushed across a surface at constant speed with a coefficient of kinetic friction of 0.24. how great a force is being used to move the crate? (use g = 9.8 n/kg and round to two significant figures.)
o 1.3 × 10² n
o 2.2 × 10³ n
o 5.3 × 10² n
o 133 n

Explanation:

Step1: Calculate the normal force

Since the crate is on a horizontal surface and there is no vertical acceleration, the normal force \(N\) equals the weight of the crate. Using the formula \(N = mg\), where \(m = 54.2\space kg\) and \(g=9.8\space N/kg\).
\(N=54.2\times9.8 = 531.16\space N\)

Step2: Calculate the frictional force

The formula for kinetic frictional force is \(F_f=\mu_kN\), where \(\mu_k = 0.24\) and \(N = 531.16\space N\) (from Step 1).
\(F_f=0.24\times531.16=127.4784\space N\)

Step3: Determine the applied force

Since the crate is moving at a constant speed, the applied force \(F_{app}\) is equal to the frictional force \(F_f\) (by Newton's first law, \(F_{net}=F_{app}-F_f = 0\) when \(a = 0\)).
Rounding \(127.4784\space N\) to two significant figures gives \(1.3\times 10^{2}\space N\)

Answer:

\(1.3\times 10^{2}\space N\)