Question

a particle of mass 100g is thrown vertically upwards with a speed of 5m/s.the work done by the force of gravity during the time particle goes up is?

Answer 1

work done =f•s
m=100g
v=5m/s

force=mass×acceleration
0.1×(-g)=-1N

since acceleration is constant so

${v}^{2} - {u}^{2} = 2as$
-25=2(-g)s
s=1.25

now work done=-1N×1.25m

-1.25 joules is answer.

Answer 2

Answer:

W.D=$1.2495J$

Step-by-step explanation:

Mass of the particle is= 100g and the initial velocity is=5m/s.

Let h be the maximum height achieved by the particle.

Using the Newton's law of motion,

$v^{2}-u^{2}=2gh$, where g is the acceleration due to gravity.

⇒$0-(5)^{2}=2(-9.8)h$

⇒$25=19.6h$

⇒$h=1.275m$

Now, The work done by the force of gravity during the time particle goes up is given by: W.D=mgh

W.D=$0.1{\times}9.8{\times}1.275$

W.D=$1.2495J$