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?

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

Expert

2022-01-11Added 26 answers

Given:

(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

And we know that

hence,

We know that the work done by some force is given by

so the work done by the cable force on the car during this process is given by

Plug from (1), and noting that cos

Now plug the given;

(b)

The work done by the gravitational force is given by

Noting that \theta= 180° since the angle between the gravitational force and the velocity is

Hence,

Noting that cos

Plug the given;

(c)

The total work done on the lift is the work done by the net force exerted on the car during this process.

Hence,

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,

Beverly Smith

Expert

2022-01-12Added 42 answers

The weight of the elevator is

$W=\left(1500kg\right)\times \left(9.8\frac{m}{{s}^{2}}\right)=1.47\times {10}^{4}N$

Because the frictional resistance is R = 100 N, the tension in the cable for dynamic equilibrium is

$T=W+R=1.47\times {10}^{4}+100=1.48\times {10}^{4}N$

By definition, the work done in lifting the elevator by 40 m is

$(1.48\times {10}^{4}N)\times \left(40m\right)=5.92\times {10}^{5}J=592kJ$

Answer:$592kJ({\textstyle \phantom{\rule{1em}{0ex}}}\text{or}{\textstyle \phantom{\rule{1em}{0ex}}}5.92\times {0}^{5}J)$

W= 1500g

R=100N

Because the frictional resistance is R = 100 N, the tension in the cable for dynamic equilibrium is

By definition, the work done in lifting the elevator by 40 m is

Answer:

W= 1500g

R=100N

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

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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