Question

A block of mass 500 g is pulled from rest on a horizontal
frictionless bench by a steady force F and travels 8 m in 2 s.
Find
a the acceleration,
b the value of F

Answer 1

Answer:

4 m/s^2 and 2 N.

Explanation:

Block was initially at rest where its velocity was 0.

When it's pulled it travels 8 m in 2 s.

  Using S = ut + (1/2)at²: Let the acceleration, here, be a.

⇒ 8 = (0)t + (1/2)a(2)²

⇒ 8 = (1/2)(4)a

⇒ 8 = 2a

⇒ 4 = a = acceleration  

Force = mass * a

            = 500 g * 4 m/s^2

            = 0.5 * 4 N

            = 2 N

Answer 2

GiVeN :-

A block of mass 500 g is pulled from rest on a horizontal  friction-less bench by a steady force F and travels 8 m in 2 s.

• Initial velocity, u = 0 m/s

• Mass, m = 500 g

• Distance, s = 8 m

• Time, t = 2 s

To FiNd :-

The acceleration (a) and the value of the steady force (F).

AcKnOwLeDgEmEnT :-

◘ Second equation of kinematics :-

$\sf{s=ut+\dfrac{1}{2}at^2 }$

◘ The force (F) acting on an object is equal to the mass (m) of an object times its acceleration (a).

F = ma

SoLuTiOn :-

(a) Putting the values :-

$\sf{\rightarrow 8=0(2)+\dfrac{1}{2}a(2)^2 }\\\\\sf{\rightarrow 8=0+\dfrac{1}{2}\times4a }\\\\\sf{\rightarrow 8=2a }\\\\\sf{\rightarrow a=\dfrac{8}{2} }\\\\\boxed{\sf{\rightarrow a=4\ m/s^2}}$

Therefore, the acceleration is 4 m/s².

$\rule{160}2$

(b) We know,

F = ma

Putting the values :-

$\sf{F = 500 \times 4}\\\\\sf{\rightarrow F = 2000\ g.m/s^2}\\\\\boxed{\sf{\rightarrow F=2\ kg.m/s^2\ or\ 2\ N}}$