Question

10 a team of dogs is pulling a sled directly south. the dogs pull with a combined force of 334 n, and the force of friction acts in the opposite direction with a magnitude of 87.0 n. if the sled weighs 91 kg, what will the speed of the sled be after 1.5 seconds? 3.7 m/s 5.5 m/s 4.1 m/s 2.7 m/s

Explanation:

Step1: Calculate the net force

The dogs pull with a force \(F_{pull}=334\ N\) south and friction \(F_{friction} = 87.0\ N\) north. The net force \(F_{net}\) is \(F_{net}=F_{pull}-F_{friction}\).

$$F_{net}=334 - 87.0=247\ N$$

Step2: Find the acceleration using Newton's second law \(F = ma\)

Given \(m = 91\ kg\) and \(F_{net}=247\ N\). Rearranging \(F = ma\) for \(a\), we get \(a=\frac{F_{net}}{m}\).

$$a=\frac{247}{91}\approx2.714\ m/s^{2}$$

Step3: Use the kinematic equation \(v = v_{0}+at\)

Assume the sled starts from rest, so \(v_{0} = 0\ m/s\), \(a\approx2.714\ m/s^{2}\), and \(t = 1.5\ s\).

$$v=0+(2.714\times1.5)\approx4.1\ m/s$$

Answer:

4.1 m/s