Question

A ball of mass 10 g is projected
upward with an initial velocity 10
m/s and comes back and finally
achieved a velocity of 5 m/s at
the point of projection. Then the
work done by air resistance.
(Neglect buoyancy force due to
air)
(A) 0.375 J
(B) 0.275 J
(C) -0.375J
(D) Zero​

Answer 1

Answer:

Option C : -0.75 J is the answer.

Explanation:

From work energy thergy theorem,

W net=W mg +W air=1/2 m(v² f+v² i),

0+W air = 1/2×0.01[(5)²-(10)²],

-0.375 J.

Hope it helps.

Answer 2

Answer:

The work done by air resistance is -0.375J.

Explanation:

Given the mass of a ball, m = 10 g

Initial velocity, $u=10~m/s$

Final velocity, $v=5~m/s$

According to the work-energy theorem, the net work done by the forces on a body is equal to the change in the kinetic energy.

i.e., $W=\Delta KE=\frac{1}{2} m(v_{f}^{2}- v_{i}^{2} )$

Here, the work done by gravity is zero, as the body is going up and coming down, $(-mg+mg=0)$.

Therefore, the net work done is only due to the air resistance,

$W=\frac{1}{2} \times0.01\times(5^{2}- (10)^{2} )$

$=0.005\times(25- 100 )$

$=0.005\times(-75)=-0.375J$

Therefore, the work done by air resistance is -0.375J.

Hence, the correct answer is option (C).