Question

question 24
1 pts
a box has a mass of 3 kilograms. the box has an initial velocity of 10 m/s. it moves across a rough surface where there is a kinetic friction coefficient equal to 0.52. calculate the distance the box will move across the ground by the time the box comes to a stop.
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Explanation:

Step1: Calculate the frictional force

The frictional force \(F_f=\mu_k N\). Since \(N = mg\) (normal force equals weight, as there is no vertical acceleration), then \(F_f=\mu_k mg\).
Substituting \(\mu_k = 0.52\), \(m = 3\space kg\), and \(g= 9.8\space m/s^2\), we get \(F_f=0.52\times3\times9.8 = 15.288\space N\).

Step2: Calculate the deceleration

Using Newton's second law \(F = ma\), so \(a=\frac{F_f}{m}\).
Substituting \(F_f = 15.288\space N\) and \(m = 3\space kg\), we get \(a=\frac{15.288}{3}=5.096\space m/s^2\) (deceleration, so \(a=- 5.096\space m/s^2\)).

Step3: Use the kinematic equation \(v^2=v_0^2 + 2ax\)

We know \(v = 0\space m/s\) (comes to a stop), \(v_0 = 10\space m/s\), and \(a=-5.096\space m/s^2\).
Rearranging the equation for \(x\), we get \(x=\frac{v^2 - v_0^2}{2a}\).
Substituting the values: \(x=\frac{0 - 10^2}{2\times(- 5.096)}=\frac{-100}{-10.192}\approx9.8\space m\).

Answer:

\(9.8\space m\)