A spring with a spring constant of 22.0 N/cm has a block attached to one end and the other end is fixed. How much work does the spring do on the block if the spring is stretched from relaxed length by 6.8 mm as the block is moved? How much additional work is done by the spring if it is stretched by an additional 6.8 mm?
To find the work done by the spring when it is stretched, we can use the formula:
Work = (1/2) k x²
Where:
- Work is the work done by the spring (in joules)
- k is the spring constant (in N/cm, which needs to be converted to N/m)
- x is the displacement or stretch of the spring (in meters)
First, let's convert the spring constant from N/cm to N/m:
1 cm = 0.01 m
So, the spring constant in N/m will be:
22.0 N/cm = 22.0 N / (0.01 m) = 2200 N/m
Now let's calculate the work done when the spring is stretched by 6.8 mm (0.0068 m):
Work = (1/2) * 2200 N/m * (0.0068 m)²
Work = (1/2) * 2200 N/m * 0.04624 m²
Work = 23.992 J (rounded to three decimal places)
So, the spring does 23.992 Joules of work on the block when it is stretched by 6.8 mm.
To find the additional work done by the spring if it is stretched by an additional 6.8 mm, we can use the same formula.
Now the total displacement or stretch of the spring will be 2 * 0.0068 m = 0.0136 m.
Therefore, the additional work done by the spring will be:
Additional Work = (1/2) * 2200 N/m * (0.0136 m)²
Additional Work = (1/2) * 2200 N/m * 0.00018496 m²
Additional Work = 0.203168 J (rounded to three decimal places)
So, if the spring is stretched by an additional 6.8 mm, it will do an additional 0.203168 Joules of work on the block.