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Rudik [331]
3 years ago
9

A 2 kg block is on a horizontal surface. A horizontal force is applied by a person to the block to pull it on the surface with a

n acceleration of 2 m/s. The coefficient of kinetic friction between the box and the surface is 0.3. How much work is done in joules by the person on the block if the block is pulled 50 m? a) 200 joules b) 250 joules c) O joules d) 300 joules e) 500 joules
Physics
1 answer:
Agata [3.3K]3 years ago
3 0

Answer:

work done = 500 J

option e is correct

Explanation:

given data

mass = 2 kg

acceleration = 2 m/s

coefficient of kinetic friction = 0.3

block D = 50 m

to find out

work

solution

we know that work done is

work done = f × d    ...........1

here f is force and d is block i.e 50 m

so here

force , f - k mg  =ma

f = 2 ( 2 + 0.3 ×10 )

f = 10 N

so from equation 1

work done =  f × d

work done = 10 × 50

work done = 500 J

option e is correct

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Answer:

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Explanation:

The magnetic force on the floating rod due to the rod held close to the ground is F = BI₁L where B = magnetic field due to rod held close the ground = μ₀I₂/2πd where μ₀ = permeability of free space = 4π × 10⁻⁷ H/m, I₂ = current in rod close to ground and d = distance between both rods = 11 mm = 0.011 m. Also, I₁ = current in floating rod and L = length of rod = 1.1 m.

So, F = BI₁L

F = (μ₀I₂/2πd)I₁L

F = μ₀I₁I₂L/2πd

Given that the current in the rods are the same, I₁ = I₂ = I

So,

F = μ₀I²L/2πd

Now, the magnetic force on the floating rod equals its weight , W = mg where m = mass of rod = 0.10kg and g = acceleration due to gravity = 9.8 m/s²

So, F = W

μ₀I²L/2πd = mg

making I subject of the formula, we have

I² = 2πdmg/μ₀L

I = √(2πdmg/μ₀L)

substituting the values of the variables into the equation, we have

I = √(2π × 0.011 m × 0.1 kg × 9.8 m/s²/[4π × 10⁻⁷ H/m × 1.1 m])

I = √(0.01078 kgm²/s²/[2 × 10⁻⁷ H/m × 1.1 m])

I = √(0.01078 kgm²/s²/[2.2 × 10⁻⁷ H])

I = √(0.0049 × 10⁷kgm²/s²H)

I = √(0.049 × 10⁶kgm²/s²H)

I = 0.22 × 10³ A

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since, WORK DONE = FORCE*DISTANCE

AND, FORCE=MASS*ACCELERATION

SO, THE WORK DONE BECOMES=MASS*ACCELERATION*DISTANCE

ACCELERATION=WORK/(MASS*DISTANCE)

AND, WORK=686J

MASS=227kg

DISTANCE=2.4m

THEREFORE, ACCELERATION=686/(227*2.4)

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The velocity time graph of an object is shown below. How far does the object travel in the time interval t =4 s to t = 6 s?
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The distance covered by the object between t =4 s and t = 6 s is 4 m

Explanation:

In a velocity-time graph, the distance covered by the object represented can be found by calculating the area under the curve.

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So, the distance covered by the object is 4 m.

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