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Brilliant_brown [7]
3 years ago
11

Two men, Joel and Jerry, each pushes an object that are identical on a horizontal frictionless floor starting from rest. Joel an

d Jerry are using the same force F. Jerry stops after 10 min, while Joel is able to push for 5.0 min longer. Compare the work they do.
Physics
1 answer:
Nataliya [291]3 years ago
7 0

Answer:

The work done by Joel is greater than the work done by Jerry.

Explanation:

Let suppose that forces are parallel or antiparallel to the direction of motion. Given that Joel and Jerry exert constant forces on the object, the definition of work can be simplified as:

W = F\cdot \Delta s

Where:

W - Work, measured in joules.

F - Force exerted on the object, measured in newtons.

\Delta s - Travelled distance by the object, measured in meters.

During the first 10 minutes, the net work exerted on the object is zero. That is:

W_{net} = W_{Joel} - W_{Jerry}

W_{net} = F\cdot \Delta s - F\cdot \Delta s

W_{net} = (F-F)\cdot \Delta s

W_{net} = 0\cdot \Delta s

W_{net} = 0\,J

In exchange, the net work in the next 5 minutes is the work done by Joel on the object:

W_{net} = W_{Joel}

W_{net} = F\cdot \Delta s

Hence, the work done by Joel is greater than the work done by Jerry.

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Answer:

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make P the subject of the equation

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(I) A novice skier, starting from rest, slides down an icy frictionless 8.0° incline whose vertical height is 105 m. How fast is
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Answer:

v = 45.37 m/s

Explanation:

Given,

angle of inclination = 8.0°

Vertical height, H  = 105 m

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Final PE = 0 J

Final KE = \dfrac{1}{2}mv^2

Using Conservation of energy

KE_i + PE_i + KE_f + PE_f

0 + m g H = \dfrac{1}{2}mv^2 + 0

v = \sqrt{2gH}

v = \sqrt{2\times 9.8 \times 105}

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A string is wrapped around a pulley with a radius of 2.0 cm. The pulley is initially at rest. A constant force of 50 N is applie
Ray Of Light [21]

Answer:

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Explanation:

Let the linear velocity of the rope(=of pulley) is v m/s

Using kinematic equation

=> v = u + at

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=>v = 4.9a ------------ eq1

By v^2 = u^2 + 2as

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=>60 = 1/2 x I x (24.49)^2

=>I = 0.20 kg-m^2

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