An object of mass 5 kg is acted upon by a force that varies with the position of the object as shown in the figure. If the object starts out from rest at the point x=0, then its speed at x=25 m is

Question

An object of mass 5 kg is acted upon by a force that varies with the position of the object as shown in the figure. If the object starts out from rest at the point x=0, then its speed at x=25 m is <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/c2c6406a75065d227569d15d5f0535c6-.png" /> (A) 5 ms -1 (B) 10 ms -1 (C) 20 ms -1 (D) 28 ms -1.

Tardigrade Admin · Accepted Answer

Till x=25 m , F=10 N ⇒ g30=(F/m r )=(10 N /5 kg )=2 ms -2 Using v2-u2=2 a s v2-02=2(2)(25), v2=100 or v=10 ms -1

Q. An object of mass $5 k g$ is acted upon by a force that varies with the position of the object as shown in the figure. If the object starts out from rest at the point $x = 0$ , then its speed at $x = 25 m$ is

$5 m s^{- 1}$

$10 m s^{- 1}$

$20 m s^{- 1}$

$28 m s^{- 1}$ .

Till $x = 25 m, F = 10 N$
$\Rightarrow g_{30} = \frac{F}{m r} = \frac{10 N}{5 k g} = 2 m s^{- 2}$
Using $v^{2} - u^{2} = 2 a s$
$v^{2} - 0^{2} = 2 (2) (25), v^{2} = 100$ or $v = 10 m s^{- 1}$

Q. An object of mass 5kg is acted upon by a force that varies with the position of the object as shown in the figure. If the object starts out from rest at the point x=0, then its speed at x=25m is

5ms−1

10ms−1

20ms−1

28ms−1.