Answer :
To solve the problem of determining how much power is needed to push a 500 lb object up a ramp with a height of 6.6 feet in 1 second, we need to go through a series of steps. Power is defined as the rate at which work is done. Here's the step-by-step process:
1. Determine the work done (W):
- The work done to lift an object vertically is given by the formula:
[tex]\[ W = F \times d \][/tex]
where [tex]\( F \)[/tex] is the force (weight of the object) and [tex]\( d \)[/tex] is the vertical distance (height).
- Here, the force [tex]\( F \)[/tex] is the weight of the object, which is 500 lb.
- The vertical distance [tex]\( d \)[/tex] is the height of 6.6 ft.
Thus:
[tex]\[ W = 500 \, \text{lb} \times 6.6 \, \text{ft} \][/tex]
[tex]\[ W = 3300 \, \text{ft-lb} \][/tex]
2. Determine the time (t):
- The time given in the problem is 1 second.
3. Calculate the power (P):
- Power is defined as the work done per unit time. The formula for power is:
[tex]\[ P = \frac{W}{t} \][/tex]
where [tex]\( W \)[/tex] is the work done and [tex]\( t \)[/tex] is the time.
- Substituting the values we have:
[tex]\[ P = \frac{3300 \, \text{ft-lb}}{1 \, \text{sec}} \][/tex]
[tex]\[ P = 3300 \, \text{ft-lb/sec} \][/tex]
Therefore, the power needed to push a 500 lb object up a ramp with a height of 6.6 feet in 1 second is [tex]\( 3300 \)[/tex] foot-pounds per second (ft-lb/sec).
1. Determine the work done (W):
- The work done to lift an object vertically is given by the formula:
[tex]\[ W = F \times d \][/tex]
where [tex]\( F \)[/tex] is the force (weight of the object) and [tex]\( d \)[/tex] is the vertical distance (height).
- Here, the force [tex]\( F \)[/tex] is the weight of the object, which is 500 lb.
- The vertical distance [tex]\( d \)[/tex] is the height of 6.6 ft.
Thus:
[tex]\[ W = 500 \, \text{lb} \times 6.6 \, \text{ft} \][/tex]
[tex]\[ W = 3300 \, \text{ft-lb} \][/tex]
2. Determine the time (t):
- The time given in the problem is 1 second.
3. Calculate the power (P):
- Power is defined as the work done per unit time. The formula for power is:
[tex]\[ P = \frac{W}{t} \][/tex]
where [tex]\( W \)[/tex] is the work done and [tex]\( t \)[/tex] is the time.
- Substituting the values we have:
[tex]\[ P = \frac{3300 \, \text{ft-lb}}{1 \, \text{sec}} \][/tex]
[tex]\[ P = 3300 \, \text{ft-lb/sec} \][/tex]
Therefore, the power needed to push a 500 lb object up a ramp with a height of 6.6 feet in 1 second is [tex]\( 3300 \)[/tex] foot-pounds per second (ft-lb/sec).
