Bruno drags a box with a mass of 50-kg along a frictionless floor by pulling the box with 18-N at an angle 40° above the horizontal. What is the net work done on the box if the displacement of the box is 10.5-m?
horizontal force= 18cos40
work done=horizontal force*distance
To find the net work done on the box, we first need to determine the force acting in the direction of displacement and the angle between that force and the displacement.
1. Resolve the applied force into horizontal and vertical components:
- The horizontal component of the force is Fcosθ, where F is the magnitude of the force (18 N) and θ is the angle (40°):
F_horizontal = F * cos(θ)
= 18 N * cos(40°)
- The vertical component of the force is Fsinθ:
F_vertical = F * sin(θ)
= 18 N * sin(40°)
2. Since the floor is frictionless, the only force acting in the horizontal direction is the horizontal component of the applied force. Therefore, the net force in the horizontal direction is equal to the horizontal component of the applied force:
F_net_horizontal = F_horizontal
3. Calculate the work done by the net horizontal force:
Work_horizontal = F_net_horizontal * displacement
= F_horizontal * displacement
= (18 N * cos(40°)) * 10.5 m
4. Since there is no vertical displacement, the vertical component of the applied force does no work:
Work_vertical = F_vertical * displacement_vertical
= 0 N * 0 m
= 0 J
5. The net work done on the box is the sum of the work done in the horizontal and vertical directions:
Net work = Work_horizontal + Work_vertical
Now you can substitute the values into the equations to get the final answer.