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Goshia [24]
3 years ago
12

While moving a box 10.0 m across the floor, Sam pushes with a force of 150 N to the right. The force friction acting on the box

is 25.0 N to the left. What is the net work done on the box by these 2 forces
Physics
1 answer:
svet-max [94.6K]3 years ago
4 0

Answer:

Explanation:required formula is

W 1=F*S

W1=work done by Sam =?

F=force applied by sam=150N

S=displacement =10m

again

W2=F*S

W2=work done by friction =?

S=displacement =10m

F=friction =25N

W=W1-W2=net work done

please feel free to ask if you have any questions

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Answer:

<h2>400 J</h2>

Explanation:

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workdone = force × distance

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workdone = 100 × 4

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<h3>400 J</h3>

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2 years ago
Does a rigid object in uniform rotation about a fixed axis satisfy the first and second conditions for equilibrium?
Vladimir [108]

Answer:

Explanation:

a rigid object in uniform rotation about a fixed axis does not satisfy both the condition of equilibrium .

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A body under uniform rotation is experiencing a centripetal force all the time so F net ≠ 0

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7 0
3 years ago
A force of 55N accelerates a 7.5kg wagon at 5.3 m/s^2 along a road. How large is the frictional force?
Varvara68 [4.7K]

Answer:

<h2>15.25 N</h2>

Explanation:

       A force of 55\text{ }N is acting on a wagon along the road. The wagon weights 7.5\text{ }kg. Acceleration of the wagon is given as 5.3\text{ }\frac{m}{s^{2}}.

       Consider the block as the system, the forces acting are Frictional force, Gravitational force, Normal reaction and External force applied by us.

       Gravitational Force and Normal Reaction cancel out each other.

       Net External Force = Mass of system/wagon \times Acceleration of wagon

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