a) Average power: 1425 W
b) Instantaneous power at 3.0 sec: 2850 W
Explanation:
a)
The motion of the object along the ramp is a uniformly accelerated motion (because the force applied is constant), so we can use the suvat equation
[tex]s=ut+\frac{1}{2}at^2[/tex]
where
s = 18 m is the displacement along the ramp
u = 0 is the initial velocity
t = 3.0 s is the time taken
a is the acceleration of the object along the ramp
Solving for a,
[tex]a=\frac{2s}{t^2}=\frac{2(18)}{(3.0)^2}=4 m/s^2[/tex]
Now we can apply Newton's second law to find the net force on the object:
[tex]F=ma=(24 kg)(4 m/s^2)=96 N[/tex]
This net force is the resultant of the applied force forward ([tex]F_a[/tex]) and the component of the weight acting backward ([tex]mg sin \theta[/tex]), so we can find what is the applied force:
[tex]F=F_a - mg sin \theta\\F_a = F+mg sin \theta = 96+(24)(9.8)(sin 37^{\circ})=237.5 N[/tex]
where
m = 24 kg is the mass of the object
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
Now we can finally find what is the work done by the applied force, which is parallel to the ramp, therefore:
[tex]W=F_a s = (237.6)(18)=4276 J[/tex]
where s = 18 m is the displacement.
Therefore the average power needed is:
[tex]P=\frac{W}{t}=\frac{4276}{3}=1425 W[/tex]
b)
The instantaneous power at any point of the motion is given by
[tex]P=F_av[/tex]
where
[tex]F_a[/tex] is the force applied
v is the velocity of the object
We already calculated the applied force:
[tex]F_a=237.5 N[/tex]
While since this is a uniformly accelerated motion, we can find the velocity at the end of the 3.0 seconds using the suvat equation:
[tex]v=u+at=0+(4)(3.0)=12.0 m/s[/tex]
And therefore, the instantaeous power at 3.0 sec is:
[tex]P=Fv=(237.5)(12)=2850 W[/tex]
Learn more about power:
brainly.com/question/7956557
#LearnwithBrainly
