Question

1. an electric motor is used to pull a 125 kg box across a floor using a long cable. the tension in the cable is 350 n and the box accelerates at 1.2 m/s² forward for 5.0 s. figure 12 the cable breaks and the box slows down and stops.

(a) calculate the coefficient of kinetic friction. ans: 0.16
(b) how far does the box travel up to the moment the cable breaks? ans: 15 m
(c) how far does the box travel from the moment the cable breaks until it stops? ans: 11 m

Explanation:

Step1: Find the frictional force

According to Newton's second law \(F_{net}=ma\). The net force \(F_{net}=T - f\) (where \(T\) is tension and \(f\) is frictional force). Given \(m = 125\space kg\), \(a=1.2\space m/s^{2}\), \(T = 350\space N\).

$$F_{net}=ma$$

$$T - f=ma$$

$$f=T - ma$$

$$f = 350-125\times1.2$$

$$f=350 - 150=200\space N$$

Step2: Calculate the normal force

On a horizontal surface, \(F_N=mg\) (where \(g = 9.8\space m/s^{2}\)). So \(F_N=125\times9.8 = 1225\space N\)

Step3: Find the coefficient of kinetic friction

The formula for frictional force is \(f=\mu_kF_N\). Rearranging for \(\mu_k\) gives \(\mu_k=\frac{f}{F_N}\)

$$ \mu_k=\frac{200}{1225}\approx0.16$$

Step4: Calculate the distance traveled before the cable breaks (for part b)

Use the kinematic equation \(d=v_0t+\frac{1}{2}at^{2}\). Since the box starts from rest \(v_0 = 0\), \(a = 1.2\space m/s^{2}\), \(t = 5.0\space s\)

$$d=\frac{1}{2}\times1.2\times5^{2}$$

$$d = 0.6\times25=15\space m$$

Step5: Find the velocity when the cable breaks (for part c)

Use \(v = v_0+at\). With \(v_0 = 0\), \(a = 1.2\space m/s^{2}\), \(t = 5.0\space s\)

$$v=0 + 1.2\times5=6\space m/s$$

After the cable breaks, the deceleration \(a'=-\frac{f}{m}=-\frac{200}{125}=- 1.6\space m/s^{2}\)
Use the kinematic equation \(v_f^{2}-v^{2}=2a'd'\). Since \(v_f = 0\)

$$0 - 6^{2}=2\times(-1.6)d'$$

$$ - 36=-3.2d'$$

$$d'=\frac{36}{3.2}=11.25\approx11\space m$$

Answer:

(a) The coefficient of kinetic friction \(\mu_k\approx0.16\)
(b) The distance traveled up to the moment the cable breaks is \(15\space m\)
(c) The distance traveled from the moment the cable breaks until it stops is \(11\space m\)