Use a fixed pulley to lift an object with a gravity of 500 N at a constant speed of 1 m / s without friction. Calculate (1) the pull force of the rope; (2) the work done by the 10 second pull force

Use a fixed pulley to lift an object with a gravity of 500 N at a constant speed of 1 m / s without friction. Calculate (1) the pull force of the rope; (2) the work done by the 10 second pull force

The fixed pulley only changes the direction of the force, not the magnitude of the force
Because the object is moving at a constant speed, the two forces balance, that is, the pull in the rope is equal to the gravity of the object
So f = g = 500N
After 10s, the moving distance of the object is s = VT = 10m, then w = FS = 5000j

When a truck with a gravity of 2.8 × 10 ^ 5N runs at a constant speed on a flat road, its resistance is 1 / 10 of its weight
Can the vehicle pass the bridge with a load limit of 25 tons

25 tons = 250000n

As shown in the figure, a pulley block is used to pull object a with a gravity of 3000n horizontally to a distance of 10m. The friction between object a and the ground is 200N, and the horizontal tension f is 120N. What are the total work, active work, additional work and the mechanical efficiency of the pulley block?

Answer:
Analysis:
1. The total work is the work done by pulling force, s = 2 × 10 = 20m
Wtotal = FS = 120 × 20 = 2400j
2. Active work is work done to overcome friction:
W = FS = 200 × 10 = 2000j
3. Extra work:
W amount = w total - W available
＝2400－2000
＝400J
4. Mechanical efficiency:
η = w yes / W total
＝2000/2400
＝83.33％

When a 2000kg vehicle runs on a horizontal highway with rated power P = 60000w, the resistance of the vehicle is 2000N
How many seconds does it take for a car to pass a distance of 300m when it reaches the maximum speed from a standstill

Resistance R = 2000N
F * Vmax = R * Vmax = P
Maximum speed: Vmax = P / r = 60000 / 2000 = 30m / S
Let the time be t,
Power work W1 = P * t,
The displacement is s = 300m, the resistance work W2 = R * s
According to the kinetic energy theorem: W1-W2 = 1 / 2m * Vmax ^ 2
60000T-2000*300=1/2*2000*30^2
T=25s