a. work=force*distance=53cos18*122
b. at constant speed, up the incline, he goes h distance upward, or 122sin6.7 high
work done=mgh=10*g*122sin6.7
b. at constant speed, up the incline, he goes h distance upward, or 122sin6.7 high
work done=mgh=10*g*122sin6.7
Work = Force * Distance * Cosine(theta)
where:
- Work is the amount of work done in joules
- Force is the amount of force applied in newtons
- Distance is the distance covered in meters
- Theta is the angle between the force and the direction of motion in degrees
Let's calculate the work done in the first scenario:
Force = 53 N
Distance = 122 m
Theta = 18 degrees
First, we need to convert the angle from degrees to radians:
Theta_radians = 18 * (pi/180) ≈ 0.31416 radians
Now, we can calculate the work done:
Work = 53 N * 122 m * Cos(0.31416) ≈ 6365.75 joules
Therefore, the man does approximately 6366 joules of work in the first scenario.
In the second scenario, where the man pulls the box up an incline, we need to consider the force exerted vertically and the force exerted parallel to the incline. The work done against the force of gravity is given by:
Force_gravity = Mass * Gravity
where:
- Mass is the mass of the box (10 kg)
- Gravity is the acceleration due to gravity (approximately 9.8 m/s^2)
Force_gravity = 10 kg * 9.8 m/s^2 = 98 N
In this case, the force applied parallel to the incline is equal to the force of gravity.
Now, let's calculate the work done:
Force_parallel = Force_gravity = 98 N
Distance = 122 m
Theta = 6.7 degrees
First, convert the angle to radians:
Theta_radians = 6.7 * (pi/180) ≈ 0.11668 radians
Work = 98 N * 122 m * Cos(0.11668) ≈ 1419.85 joules
Therefore, the man does approximately 1420 joules of work in the second scenario.