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question 12
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 work done by the gravitational force?
1,760 j
1,390 j
98.0 j
0 j

Explanation:

Step1: Recall the work formula

The work formula is \(W = Fd\cos\theta\), where \(F\) is the force, \(d\) is the displacement, and \(\theta\) is the angle between the force and displacement vectors.

Step2: Analyze the gravitational force and displacement

The gravitational force \(F_g=mg\) acts vertically downward (\(y -\) direction). The sled moves horizontally (\(x -\) direction). So, the angle \(\theta = 90^{\circ}\) between the gravitational force and the displacement of the sled.

Step3: Calculate the work done by gravitational force

Substitute \(\theta = 90^{\circ}\) into the work formula. Since \(\cos(90^{\circ})=0\), then \(W = Fd\cos(90^{\circ})=0\)

Answer:

\(0J\)