How Is Work Calculated When Lifting and Carrying a Cement Block?

[hshphyss]

A person lifts a 5.5 kg cement block a vertical distance 1.2 m and then carries the block horizontally a distance 7.3 m. Determine the work done by the person and by the force of gravity in this process. How much work is done by the person and the force of gravity?

Would you break this up into components?

 

[whozum]

Its not necessary if you recall the definition of work to involve a dot product.

 

[quicknote]

Yes, it would be helpful to break this problem into components in order to accurately calculate the work done by the person and the force of gravity. Let's first define the variables involved in this problem:

m = mass of the cement block (5.5 kg)
d1 = vertical distance lifted (1.2 m)
d2 = horizontal distance carried (7.3 m)
g = acceleration due to gravity (9.8 m/s²)

To calculate the work done by the person, we can use the equation W = Fd, where W is work, F is the force applied, and d is the displacement. In this case, the force applied by the person is equal to the weight of the block, which can be calculated using the formula F = mg. Therefore, the work done by the person is:

W = (mg)(d1 + d2)
= (5.5 kg)(9.8 m/s²)(1.2 m + 7.3 m)
= 437.45 joules

To calculate the work done by the force of gravity, we can use the same equation, but this time the displacement is only in the vertical direction (d1). The force of gravity is also acting in the opposite direction of the displacement, so we will use a negative sign to indicate this direction. Therefore, the work done by the force of gravity is:

W = -(mg)(d1)
= -(5.5 kg)(9.8 m/s²)(1.2 m)
= -64.68 joules

In summary, the work done by the person is 437.45 joules and the work done by the force of gravity is -64.68 joules. The negative sign indicates that the force of gravity is doing negative work, or work in the opposite direction of the displacement. This is because the block is being lifted against the force of gravity, so the force and displacement are in opposite directions.