Question

A crane on a building site lifts bricks of total mass 200 kg, initially at rest on the ground, with uniform acceleration. When the bricks are 5.0m from the ground, they have a speed of 5.0 m / s. How much work is done during this process?

Answer 1

Answer:

w=force/ time.f=mg.m=mass,g=acceleration due to gravity.

now,m is 200 so f is 200×10=2000 neuton.

now speed=distance/time so,time=distance/speed=(5m)/(5m/s)

=1sec.

here,w=f/t so,

2000/1=2000joule

Answer 2

Answer:

The work done during this process is 12300 J.

Explanation:

Given that,

Mass = 200 kg

Height = 5.0 m

speed =5.0 m/s

The mechanical energy is defined as,

$E=\dfrac{1}{2}mv^2+mgh$

Where, m = mass

g = acceleration due to gravity

h = height

v = speed

Put the value into the formula,

$E =\dfrac{1}{2}\times200\times(5)^2+ 200\times9.8\times5$

$E=12300\ J$

The work done is change in energy.

$W = \Delta E$

$W =\dfrac{1}{2}mv^2+mgh-0$

$W=12300\ J$

Hence, The work done during this process is 12300 J.