Respuesta :
Answer:
the tractor's power output = 1318.47 watts
Explanation:
Weight of bale = (mg) = (150 kg * 9.81 m/s²) = 1471.5 newtons
Resolve that weight force into its components at right angles to the ramp and parallel to the slope.
Down the slope we have 1471.5sin(15°) = 380.85 N
At right angles to the ramp we have 1471.5cos(15°) = 1421.35 N
OK, now we need the relationship between the coefficient of friction (μ), the friction force (Ff) and the normal force (Fn) {The normal force is the force at right angles to the friction surface.  In this case Fn is equal in magnitude to the component of weight force at right angles to the surface.}
Ff = μ * Fn = (0.40 * 1421.35) = 568.54 N
The bale is not accelerating, so the "pull force" up the incline = component of weight + friction force down
= (380 N + 568.54 N)
= 948.54N
We need the speed in m/s not km/h
5.0 km/h = 5000 m/h
= (5000/3600)
= 1.39 m/s
Power = (948.54 N * 1.39m/s)
= 1318.47 N.m/s
= 1318.47 watts
The power output of the tractor is 1318.47 W
Power:
According to the question the weight of bale
mg = 150 × 9.81 = 1471.5 N
Resolving the weight we get:
The normal reaction perpendicular to the ramp is given by:
N = 1471.5cos(15°) = 1421.35 N
The force of friction is given by:
[tex]f = \mu N = 0.40 \times 1421.35\\\\f = 568.54 N[/tex]
The bale is not accelerating, so the force up the incline = component of weight + friction force down
F = mgsin(15°) + f
F = 380 N + 568.54 N
F = 948.54N
Now, the engine speed is:
v = 5.0 km/h = 5000 m/h
v = (5000/3600)
v = 1.39 m/s
The power is defined as the product of force and speed, so:
Power P = Fv
P = 948.54 × 1.39
P = 1318.47 W
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