Question

problem reference 6.1
a boy pulls a 10.0 kg sled at a constant velocity for a distance of 18.0 m. he pulls in a direction 38.0° above the horizontal. a 24.0 n frictional force opposes the motion of the sled.
what is the magnitude of the force that the boy exerts on the sled?
24.0 n
39.0 n
30.5 n
56.7 n

Explanation:

Step1: Analyze the horizontal forces

Since the sled moves at a constant velocity, the net force in the horizontal direction is zero. Let the force exerted by the boy be \(F\). The horizontal component of \(F\) is \(F\cos\theta\), and the frictional force \(f = 24.0\space N\). So, \(F\cos\theta=f\).

Step2: Solve for \(F\)

We know \(\theta = 38.0^{\circ}\) and \(f = 24.0\space N\). Then \(F=\frac{f}{\cos\theta}\). Substituting the values, \(\cos(38.0^{\circ})\approx0.788\), so \(F=\frac{24.0}{0.788}\approx30.5\space N\).

Answer:

\(30.5\space N\)