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iren [92.7K]
2 years ago
9

A block with a mass of 2kg is pushed 20cm and a constant speed up an incline that makes a 40 degree angle with the floor. how mu

ch work is done by gravity on the block ? let d be the displacement of the block.
Physics
1 answer:
Alexxx [7]2 years ago
8 0

work done will be 2.5872 J

Given :

mass of block = 2kg

distance travelled by block = 20cm

angle of inclination = 40°

To Find :

Work done by gravity

Solution :

Gravity is defined as the force that attracts a body towards the earth or towards any other physical body having mass.

work done by gravity is mgh

If θ is the angle made when the body falls, the work done by gravity is given by,

W = m g h cosθ

W = 2 x 9.8 x 0.2 x cos40°

= 2.5872 J

So work done will be 2.5872 J

Learn more about Work done here:

brainly.com/question/25923373

#SPJ4

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Now the elevator is moving downward with a velocity of v = -2.8 m/s but accelerating upward with an acceleration of a = 5.5 m/s2
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160.75 N

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An air hockey game has a puck of mass 30 grams and a diameter of 100 mm. The air film under the puck is 0.1 mm thick. Calculate
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Answer:

time required after impact for a puck is 2.18 seconds

Explanation:

given data

mass = 30 g = 0.03 kg

diameter = 100 mm = 0.1 m

thick = 0.1 mm = 1 ×10^{-4} m

dynamic viscosity = 1.75 ×10^{-5} Ns/m²

air temperature = 15°C

to find out

time required after impact for a puck to lose 10%

solution

we know velocity varies here 0 to v

we consider here initial velocity = v

so final velocity = 0.9v

so change in velocity is du = v

and clearance dy = h

and shear stress acting on surface is here express as

= µ \frac{du}{dy}

so

= µ  \frac{v}{h}   ............1

put here value

= 1.75×10^{-5} × \frac{v}{10^{-4}}

= 0.175 v

and

area between air and puck is given by

Area = \frac{\pi }{4} d^{2}

area  =  \frac{\pi }{4} 0.1^{2}

area = 7.85 × \frac{v}{10^{-3}} m²

so

force on puck is express as

Force = × area

force = 0.175 v × 7.85 × 10^{-3}

force = 1.374 × 10^{-3} v    

and now apply newton second law

force = mass × acceleration

- force = mass \frac{dv}{dt}

- 1.374 × 10^{-3} v = 0.03 \frac{0.9v - v }{t}

t =  \frac{0.1 v * 0.03}{1.37*10^{-3} v}

time = 2.18

so time required after impact for a puck is 2.18 seconds

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