Question

Velocity-time graph of a particle moving in a straight line is as shown in figure. Mass of the particle is 2kg. Work done by all the forces acting on the particle in time interval between t= 0 to t= 10 Seconds.
is?
la) 300J
(b)-300 J
(c) 400 J
(d)-400 J


Pls send. ​

Answer 1

Answer:

300J

Explanation:

work done= change in kinetic energy.

= 1/2m(v^2-u^2)

=1/2 x 2 x [(20^2)-(10^2)]

=1 x [400-100]

=300J

If you need me to elaborate, let me know in the comment section.

Answer 2

Answer:

• Work done (W) is 300 Joules.

Given:

1. Mass of The particle (M) = 2 kg.
2. Time interval ; t = 0 Seconds to t = 2 Seconds.

Explanation:

$\rule{300}{1.5}$

From Work - Energy Theorem We Know,

$\large \bigstar \; \boxed{\tt W = k_f - k_i} \\\\\\ \bold{Here}\begin{cases}\text{W Denotes Work Done} \\ \sf{K_i} \text{ Denotes Initial Kinetic energy} \\ \sf{K_f} \text{ Denotes Final Kinetic energy}\end{cases}$

Now,

$\boxed{\tt W = K_f - K_i}$

Substituting the Values.

$\tt \longmapsto W = \dfrac{1}{2} \times M v^2 - \dfrac{1}{2} \times M u^2 \\\\\\ \longmapsto W = \dfrac{1}{2} \times M ( v^2 - u^2) \\\\\\ \longmapsto W = \dfrac{1}{2} \times M ( [- 20]^2 - [10]^2) \\\\\\ \longmapsto W = \dfrac{1}{2} M \times ( 400 - 100) \\\\\\ \longmapsto W = \dfrac{1}{2} \times M \times 300 \\\\\\ \bullet Mass = 2 \; Kg \\\\\\ \longmapsto W = \dfrac{1}{2} \times 2 \times 300 \\\\\\ \longmapsto W = \dfrac{1}{\cancel{2}} \times \cancel{2} \times 300 \\\\\\ \longmapsto W = 1 \times 300$

$\longmapsto {\underline{\boxed{\red{\tt W = 300 \; J}}}}$

∴ Work done (W) on the Particle by all the Force is 300 Joules.

Note:-

• Initial Velocity And Final Velocity has been Taken According to the Graph given.  

$\rule{300}{1.5}$