a) Calculate the work done on a 1500-kg elevator car by its cable to lift...

Ernest Ryland

Ernest Ryland

Answered

2022-01-10

a) Calculate the work done on a 1500-kg elevator car by its cable to lift it 40.0 m at constant speed, assuming friction averages 100 N.
b) What is the work done on the lift by the gravitational force in this process?
c) What is the total work done on the lift?

Answer & Explanation

GaceCoect5v

GaceCoect5v

Expert

2022-01-11Added 26 answers

Given:
mcar=1500kg
y=40m
Fs=100N
(a)
First of all, we need to find the force of the cable.
Noting that the friction force always working in the opposite direction of the motion of the object.
We know, from Newton's second law, that
Fy=FcablemgFs=may
And we know that ay=0 since the velocity is constant.
hence,
FcablemgFs=0
Fcable=mg+Fs
We know that the work done by some force is given by
W=Fdcosθ
so the work done by the cable force on the car during this process is given by
Wcable=Fcableycos0
Plug from (1), and noting that cos 0=1.0
Wcable=(mg+Fs)y
Now plug the given;
Wcable=[(1500x9.8)+100]×40
Wcable=5.92x10J
(b)
The work done by the gravitational force is given by
Wcable=Fgdcosθ
Wcable=mgycosθ
Noting that \theta= 180° since the angle between the gravitational force and the velocity is 180
Hence,
Wgravity=mgycos180
Noting that cos 180=1
Wgravity=mgy
Plug the given;
Wgravity=1500×9.8×40
Wgravity=5.88×105J
(c)
The total work done on the lift is the work done by the net force exerted on the car during this process.
Hence,
Wtot=Fnetθy
We know that the net force exerted on the car during this process is zero since the car is moving at a constant speed (see part a above).
So,
Wtot=0×40
Wtot=0J
image

Beverly Smith

Beverly Smith

Expert

2022-01-12Added 42 answers

The weight of the elevator is
W=(1500kg)×(9.8ms2)=1.47×104N
Because the frictional resistance is R = 100 N, the tension in the cable for dynamic equilibrium is
T=W+R=1.47×104+100=1.48×104N
By definition, the work done in lifting the elevator by 40 m is
(1.48×104N)×(40m)=5.92×105J=592kJ
Answer:592kJ(or5.92×05J)
W= 1500g
R=100N
star233

star233

Expert

2022-01-15Added 238 answers

Given
mass of elevator m=1500 kg
Lifting height =40 m
Friction Force F=100 N
The cable has to work against gravitational force and friction Force
Increase in Potential Energy of Elevator =mgh=1500×9.8×40=588 kJ
thus work done by gravitational force = 588 kJ
Work done against friction force = F * x =100 * 40=4 kJ
Total Work done by the cable = 592 kJ

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