Final answer:
The rate at which work is done, or power, in lifting a 35-kilogram object at a constant speed of 5.0 meters per second is 1715 Watts.
Step-by-step explanation:
The rate at which work is done in lifting a 35-kilogram object vertically at a constant speed of 5.0 meters per second is known as the power output during lifting. The formula to calculate work done is work (W) = force (F) × distance (d). However, since the object is lifted at a constant speed, the force applied is equal to the weight of the object, which is the mass times the acceleration due to gravity (F = m × g, where g=9.8 m/s2). Therefore, the work done per second, or power (P), is going to be P = F × v, where F is the force and v is velocity.
Calculating this:
- Force (F) = mass (m) × gravity (g) = 35 kg × 9.8 m/s2 = 343 N
- Power (P) = F × velocity (v) = 343 N × 5.0 m/s = 1715 Watts
Thus, the power output is 1715 Watts.