Question

A force'F' act on abody of mass m initially at rest producing a uniform acceleration ,'a' for a time interval 't'
Then the work done W on the body is

Answer 1

Answer:

1/2 m a^2 t^2

Explanation:

acceleration = a

time = t

mass = m

Therefore,

Work = F × d

= ma × (ut + 1/2at^2). {d= ut+1/2at^2}

= ma × (0 + 1/2 at^2). since u=0

= ma× 1/2at^2

= 1/2 m a^2 t^2.

Answer 2

Answer:

$\frac{1}{2} ma^{2}t^{2}$

Explanation:

Given:

A force (F) acting on a mass (m) body causes a uniform acceleration (a) for a period of time (t).

To find:

Work done

Solution:

Work in mathematics is displacement multiplied by force.

W= F x d

W stands for work, F represents the force's strength, and d stands for displacement.

A force is considered to do positive work when it is applied if it contributes to the displacement of the site of application. A force performs negative work if one of its components is at the application point displaced in the opposite direction.

The SI unit of work, the joule (J), is defined as the amount of effort required to exert one newton of force across a one-meter distance.

Non-SI units of work include the Newton-metre, erg, foot-pound, kilowatt-hour, litre-atmosphere, and horsepower-hour.

a = acceleration, time = t, mass = m

Therefore,

Work = F×d

$=ma(ut+\frac{1}{2}at^{2})$

Because we know that

$d=ut+\frac{1}{2}at^{2}$

Here since the value of u = 0

$=ma(0+\frac{1}{2} at^{2} )\\=(ma)(\frac{1}{2} at^{2} )\\=\frac{1}{2}a^{2}t^{2}$

Calculate the work done by the effort when

a load of 60kgf is lifted through a distance of

20m by applying an effort of 30kgf which is

displaced through 60m using a pulley having

velocity ratio of 3.

https://brainly.in/question/47998975

0.5 KJ of work is done by a person who uses a force of 50N to move a box.How far does it go?​

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