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sladkih [1.3K]
2 years ago
13

A fully loaded, slow-moving freight elevator has a cab with a total mass of 1250 kg, which is required to travel upward 49 m in

3.1 min, starting and ending at rest. The elevator's counterweight has a mass of only 995 kg, and so the elevator motor must help. What average power is required of the force the motor exerts on the cab via the cable?
Physics
1 answer:
insens350 [35]2 years ago
5 0

Answer:

659.01W

Explanation:

The cab has a mass of 1250 kg, the weight of the cab represented by Wc will be

Wc = mass of the cab × acceleration due to gravity in m/s²

Wc = 1250 × 9.81 = 12262.5 N

but the counter weight of the elevator represented by We = mass × acceleration due to gravity = 995 × 9.81 = 9760.95 N

Net weight = weight of the cab - counter weight of the elevator = Wc - We = 12262.5 - 9760.95 = 2501.55 N

the motor of the elevator will have to provide this in form of work

work done by the elevator to lift the cab to height of 49 m = net weight × distance (height) = 2501.55 × 49m

power provided by the motor of the elevator = workdone by the motor / time in seconds

Power = (2501.55 × 49) ÷ ( 3.1 × 60 seconds) = 659.01 W

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▪▪▪▪▪▪▪▪▪▪▪▪▪  {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪

Let's solve ~

Given terms :

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  • velocity (v)= 4 m/s

The formula to find kinetic Energy is ~

\boxed{ \boxed{ \sf{ \frac{1}{2}  m{v}^{2} }}}

Now, apply the formula according to given situation

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \: \dfrac{1}{2}  \times 7 \times ( {4)}^{2}

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \: \dfrac{1}{2}  \times 7 \times 16

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:7 \times 8

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:56 \:  \: joules

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