Question

1. a snowmobile is used to pull two sleds across the ice. the mass of the snowmobile and the rider is 320 kg. the mass of the first sled behind the snowmobile is 120 kg and the mass of the second sled is 140 kg. the ground exerts a force of 1500 n forward on the snowmobile. the coefficient of kinetic friction for the sleds on ice is 0.15. assume that no other frictional forces act on the snowmobile. calculate the acceleration of the snowmobile and sleds. ans: 1.9 m/s² forward

Explanation:

Step1: Calculate the total mass

The total mass \(m_{total}\) is the sum of the mass of the snowmobile and rider (\(m_1 = 320\space kg\)), the mass of the first sled (\(m_2=120\space kg\)) and the mass of the second sled (\(m_3 = 140\space kg\)).
\(m_{total}=m_1 + m_2+m_3=320 + 120+140=580\space kg\)

Step2: Calculate the frictional force on the sleds

The normal force on the sleds is equal to their weight. The weight of the sleds \(F_g=(m_2 + m_3)g\), where \(g = 9.8\space m/s^2\). The frictional force \(F_f=\mu_kF_g\), with \(\mu_k = 0.15\).
\(F_g=(120 + 140)\times9.8=260\times9.8 = 2548\space N\)
\(F_f=\mu_kF_g=0.15\times2548 = 382.2\space N\)

Step3: Apply Newton's second law

According to Newton's second law \(F_{net}=F_{applied}-F_f=m_{total}a\). We know \(F_{applied}=1500\space N\) and \(F_f = 382.2\space N\), \(m_{total}=580\space kg\).
\(a=\frac{F_{applied}-F_f}{m_{total}}=\frac{1500 - 382.2}{580}=\frac{1117.8}{580}\approx1.9\space m/s^2\)

Answer:

The acceleration of the snowmobile and sleds is \(1.9\space m/s^2\) [forward].