A pickup truck is hauling a barge along a canal at a constant speed. The truck, driving parallel to the waterway, is attached to the barge by a cable tied to the bow making a 29.8 ∘ angle with the forward direction. If the truck exerts a force of 870 N on the cable, how much work is done in overcoming friction as the barge is moved 10.70 km.
Answers
Answer:
The work done W is defined as the product of the force F acting on a body and the displacement s of the body with a cosine angle θ. along the horizontal direction, the distance completed by the body is similar to the displacement s.
The expression of work done W in terms of force F and displacement s is
Explanation:
A pickup truck is hauling a barge along a canal at a constant speed. The truck, driving parallel to the waterway, is attached to the barge by a cable tied to the bow making a 29.8 ∘ angle with the forward direction.
If the truck exerts a force of 870 N on the cable , how much the work is done in overcoming friction as the barged is moved 10.7 km ?
work done of an object is defined as the product of the force applied to the forceand the displacement of the object along the direction of force.
it is expressed as , W = FscosФ , where F is applied force, s is displacement and Ф is angle between displacement and force.
here, F = 870 N , s = 10.7 km = 10700 m and Ф = 29.8° ≈ 30°
∴ W = 870 N × 10700 m × cos30°
= 8.7 × 10.7 × √3/2 × 10⁵
= 8.062 × 10⁶ J