To push a 33.0-kg crate up a frictionless incline angled at 27.0° to the horizontal, a worker exerts a force of 388 N parallel to the incline. The crate moves a distance of 4.10 m.
What work is done on the crate by the worker? What work is done on the crate by the weight of the crate? What work is done by the normal force exerted by the floor on the crate?What is the total work done on the crate?
To find the work done on the crate by the worker, we can use the equation:
Work = Force * Distance * cos(theta)
Where:
- Work is the work done on the crate
- Force is the force exerted by the worker (388 N)
- Distance is the distance the crate moves (4.10 m)
- theta is the angle between the force and the direction of motion (27.0°)
So, the work done on the crate by the worker is:
Work = 388 N * 4.10 m * cos(27.0°)
To find the work done on the crate by the weight of the crate, we can use the equation:
Work = Weight * Distance * cos(theta)
Where:
- Work is the work done on the crate by the weight of the crate
- Weight is the weight of the crate, which is equal to its mass multiplied by the acceleration due to gravity (9.8 m/s^2)
- Distance is the distance the crate moves (4.10 m)
- theta is the angle between the weight and the direction of motion (27.0°)
So, the work done on the crate by the weight of the crate is:
Work = (mass * acceleration due to gravity) * 4.10 m * cos(27.0°)
To find the work done by the normal force exerted by the floor on the crate, we need to note that the normal force only acts perpendicular to the direction of motion. Since the crate moves parallel to the incline, there is no work done by the normal force in this case.
Finally, the total work done on the crate is the sum of the work done by the worker and the work done by the weight of the crate:
Total Work = Work by worker + Work by weight of crate
I'll calculate the values for you.