Force needed to push a box off a cliff

[paigegail]

HELP! I am in grade 12 physics and my teacher gave us a puzzler for the weekend. I need serious help with it and all of my classmates can't figure it out either! The question is :

A 10.8 kg box is 5.0 m from the edge of a 12.96 m high cliff. Bob pushes the box over the edge of the cliff by applying a constant horizontal force to it until the box goes over the edge. The force of friction on the box is 49.0 N and the box lands 7.5 m from the base of the cliff. How much force did Bob apply to the box while he was pushing?

What do I do and can someone show me how to solve it?

 

[mani]

A 10.8 kg box is 5.0 m from the edge of a 12.96 m high cliff. Bob pushes the box over the edge of the cliff by applying a constant horizontal force to it until the box goes over the edge. The force of friction on the box is 49.0 N and the box lands 7.5 m from the base of the cliff. How much force did Bob apply to the box while he was pushing?

If v is the (horiz.) velocity at push-off and t the time of fall;
vt = 7.5; (1/2) gt^2 = 12.96; Find v.
Total work done by all forces till push-off,
(F-49)5 = Kinetic energy at push-off = (1/2) (10.8) v^2

 

[M98Ranger]

First of all, don't panic! Puzzles like this can be intimidating at first, but with some careful thinking and a few key equations, you'll be able to solve it. Let's break down the problem and go through the steps to find the solution.

Step 1: Draw a diagram. This will help you visualize the problem and identify all the given values and unknowns. In this case, draw a cliff with the box on top and label the distance from the edge of the cliff (5.0 m), the height of the cliff (12.96 m), and the distance the box lands from the base of the cliff (7.5 m).

Step 2: Identify the forces acting on the box. In this case, we have the force of gravity pulling the box down, the normal force from the cliff pushing the box up, and the force of friction opposing the motion of the box.

Step 3: Write down the equations that relate the forces to the motion of the box. In this case, we can use Newton's second law, F=ma, to relate the net force on the box to its acceleration. We can also use the equation for work, W=Fd, to relate the force applied by Bob to the displacement of the box.

Step 4: Set up a system of equations. We know that the net force on the box is equal to the mass of the box times its acceleration, so we can write: ΣF=ma. We also know that the work done by Bob is equal to the force he applied multiplied by the displacement of the box, so we can write: W=Fd. We can also use the equation for work to find the force of friction, since we know the displacement (7.5 m) and the work done (49.0 Nm). So we can write: W=Fd=49.0 Nm.

Step 5: Solve the equations. We now have three equations with three unknowns: the force applied by Bob (F), the acceleration of the box (a), and the force of friction (Ff). We can solve for these variables by plugging in the given values and using algebra to rearrange the equations.

Step 6: Check your units and answer. Make sure all your units are consistent and that your final answer is in Newtons (N), which is the unit for force. You should find that the force applied by Bob is