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This guide explores the concept of work in physics, focusing on the relationship between force, distance, and angle. Work is defined as the product of the force applied to an object in the direction of its displacement. Key principles such as scalar quantities, angles affecting work done, and the dot product are discussed. Practical examples, including scenarios involving lifting and carrying objects, illustrate how to calculate work using formulas like W = F · D cos(θ). Solve problems involving forces acting on objects to strengthen understanding of this essential physics concept.
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Total Recall • Last year, you learned • Work = Force x Distance • Work is measured in Joules (J) • Work is a scalar quantity
Work is only done on an object if the Force F is in the same ‘plane’ as the object’s displacement Lift books – Work done? Yes F D Carry books – Work done? No F D
For an object with a force being applied at an angle, this represents a unique situation • Not all of the force is going to be in the same ‘plane’ as the displacement
To find the Work done, we would need to know Fx (assuming the object was moving to the right) • Technically, we would need to subtract the friction too!
So…..given this diagram We can find the Fx from Fx = Fcos θ Fx = 150 cos 60 = 75 N And so W = 75 N x 25 m = 1875 J 25 m
So?? • Well let’s look at that problem again…. We found W = (Fcos θ)D Rearrange that…. W = FD cos θ Look familiar?? Replace F and D with a and b
W = FDcos θ also means that W = F·D • This is only true for a constant force
The first two examples of work also makes sense Lift books – Work done? Yes angle is 0o, so i·i, j·j, k·k which all = 1 Carry books – Work done? No angle is 90o, so i·j, j·k, i·k which all = 0
PROBLEMS!!!!!! • A floating ice block is pushed through a displacement of 15i – 12j m along a straight embankment by rushing water, which exerts a force F = 210i – 150j N on the block. How much work was done on the block?
A shopper in a supermarket pushes a cart with a force of 35 N directed at an angle of 25o downward from the horizontal. Find the work done by the shopper as she moves down the aisle 50 m in length.
A particle moving in the xy plane undergoes a displacement d= 2i + 3j m as a constant force F = 5i + 2j N acts on the particle. • a. Calculate the magnitude of the force and the displacement. • b. Find the work done on the particle. • c. Find the angle between the Force and the Distance traveled.
A force F = 6i – 2j N acts on a particle that undergoes a displacement from a point d1 = 4i - 6j m to a point d2 = -i +7j m. At what angle is the force acting on the object relative to it’s displacement?