Question

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

Answer 1

Answer:

Explanation:

As we learnt in:

Net work done by all the forces give the change in kinetic energy -

W=\frac{1}{2}mv^{2}-\frac{1}{2}mv{_{0}}^{2}

W= k_{f}-k_{i}

- wherein

m=mass \: of\: the\: body

v_{0}= initial\: velocity

v= final\: velocity

 

 

Work done by force of gravity is change in kinetic energy while moving up .

\therefore\ W_{g}=\frac{1}{2}mv_{f}^{2}-\frac{1}{2}mv_{i}^{2}

   W_{g}=0-\frac{1}{2}\times (0.1kg)(25)

   W_{g}=-1.25J

Answer 2

Answer:

-mgH

Where H = maximum height=u^2/2g

Work done by gravity = -0.1kg×g×5^2/2g

= -1.25 joules

Explanation: