Question

1. a 75 kg bobsled is pushed along a horizontal surface by two athletes. after the bobsled is pushed a distance of 4.5 m starting from rest, its speed is 6.0 m/s. find the magnitude of the net force on the bobsled.

m = 75 kg
x = 4.5 m
v = 6.0 m/s
v₀ = 0 m/s

w = f·d
kef - kei = f·d
f = (kef - kei)/d = (½mv²)/d

Explanation:

Step1: Recall Work - Energy Theorem

The work - energy theorem states that the work done on an object \(W\) is equal to the change in its kinetic energy \(\Delta KE\). The formula for work is \(W = F\cdot d\) (where \(F\) is the net force and \(d\) is the distance), and the formula for kinetic energy is \(KE=\frac{1}{2}mv^{2}\). The initial velocity \(v_{0} = 0\ m/s\), so the initial kinetic energy \(KE_{i}=\frac{1}{2}mv_{0}^{2}=0\ J\). The final kinetic energy \(KE_{f}=\frac{1}{2}mv^{2}\), where \(m = 75\ kg\) and \(v=6.0\ m/s\), and the distance \(d = 4.5\ m\).

Step2: Calculate Final Kinetic Energy

First, calculate the final kinetic energy. Using the formula \(KE_{f}=\frac{1}{2}mv^{2}\), substitute \(m = 75\ kg\) and \(v = 6.0\ m/s\):

$$KE_{f}=\frac{1}{2}\times75\ kg\times(6.0\ m/s)^{2}=\frac{1}{2}\times75\times36\ J= 75\times18\ J = 1350\ J$$

Step3: Use Work - Energy Theorem to Find Force

From the work - energy theorem, \(W=\Delta KE=KE_{f}-KE_{i}\). Since \(KE_{i} = 0\), \(W = KE_{f}\). Also, \(W=F\cdot d\), so \(F\cdot d=KE_{f}\). We can solve for \(F\) by rearranging the formula: \(F=\frac{KE_{f}}{d}\)
Substitute \(KE_{f}=1350\ J\) and \(d = 4.5\ m\) into the formula:

$$F=\frac{1350\ J}{4.5\ m}=300\ N$$

Answer:

The magnitude of the net force on the bobsled is \(\boldsymbol{300\ N}\)