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mel-nik [20]
4 years ago
8

(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 averag

es 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?
Physics
1 answer:
Ahat [919]4 years ago
8 0

Answer:592 kJ

Explanation:

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=mg h=1500\times 9.8\times 40=588 kJ

thus work done by gravitational force=588 kJ

Work done against friction force=F\cdot x=100\cdot 40=4 kJ

Total Work done by the cable=592 kJ

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Answer:

J = 12.32kg*m/s

Explanation:

Assumptions: I'm assuming mis is m/s

Given: The baseball's mass is 140g, so convert to kg, 0.140kg. (Good rule of thumb, in physics convert grams to kilograms). It is initially traveling 35 m/s to the right.

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Now the equation for impulse is (dp is Δp, which is difference in momentum; dv is Δv, the difference in velocity; J is impulse):

J = dp= m*dv

To get Δv, we have to find the difference of velocity, that is why we broke it into components. I'm going to define right as positive and left as negative. After that, we find the velocity vector:

dv_{x} = (35-(-49.84692829)) = 84.84692829 m/s\\dv_{y} = (0-(23.2440044)) = -23.2440044 m/s\\dv = \sqrt{(84.84692829)^2+(-23.2440044)^2} = 87.97320604m/s

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J = 12.32kg*m/s

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