Question

(problem reference m.6)
a boy pulls a 5.0 - kg sled with a rope that makes a 60.0° angle with respect to the horizontal surface of a frozen pond. the boy pulls on the rope with a force of 10.0 n; and the sled moves with constant velocity.
what is the magnitude of the kinetic frictional force acting between the sled and the ice?
49 n
10 n
5.0 n
8.7 n
40 n

Explanation:

Step1: Analyze the forces in the horizontal direction

Since the sled moves with constant velocity, the net force in the horizontal direction is zero. The horizontal component of the pulling force \(F\) is \(F\cos\theta\), and this must balance the kinetic frictional force \(f_k\).
The formula for the horizontal component of the force is \(F_x = F\cos\theta\), where \(F = 10.0\ N\) and \(\theta=60.0^{\circ}\)

Step2: Calculate the horizontal component of the force

Substitute the values into the formula:
\(F_x=10.0\cos60.0^{\circ}\)
We know that \(\cos60.0^{\circ}=\frac{1}{2}\)
So \(F_x = 10.0\times\frac{1}{2}=5.0\ N\)

Answer:

5.0 N