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

GIVEN

• Mass of block = 500 g
• Initially at rest
• Distance travelled = 8 m
• Time = 2 seconds.

TO FIND

• a) Acceleration of block
• b) Value of force

SOLUTION

We have,

• Distance travelled (s) = 8 m

• Initial speed (u) = 0

• Time (t) = 2 s

• Mass (M) = 500 g

We will use 2nd equation of motion :

$\: \dag \: \: \boxed{\boxed{\tt\blue{s = ut + \dfrac{1}{2}\: at^2 }}} \\\\ \qquad\underline{\underline{\rm{Substituting \: values }}} \\\\ \blue{\longmapsto{\sf{8 = 0 \times 2 + 1/2 \: \times a \times 2^2 }}} \\\\ \red{\longmapsto{\sf{8 = 0 + 1/2 \times a \times 4 }}} \\\\ \green{\longmapsto{\sf{8 = 2a }}} \\\\ \red{\longmapsto{\sf{a = 8/2 }}} \\\\ \blue{\longmapsto{\boxed{\boxed{\sf\purple{a = 4 \: ms^{-2} }}}}} \\\\ \therefore\underline{\sf\red{a) Acceleration \: of \: block = 4 \: m/s^2 }}$

$\rule{300}{1}$

b) We now have,

• Mass of block (M) = 500 g = 0.5 kg

• Acceleration of block (a) = 4 m/s²

Now we will use 2nd law :

$\: \dag \: \: \boxed{\boxed{\tt\blue{Force = Mass \times acceleration }}} \\\\ \qquad\underline{\underline{\rm{Substituting \: values }}} \\\\ \red{\longmapsto{\sf{Force = 0.5 \times 4 }}} \\\\ \blue{\longmapsto{\boxed{\boxed{\sf\blue{Force = 2 \: kgm/s^2 \: or \: 2\: N }}}}} \\\\ \therefore\underline{\sf\green{ b) Force \: exerted \: by \: block \: = 2 \: N }}$

Answer 2

GIVEN:

• Mass of block = 500g = 0.5 kg
• Distance travelled = 8m
• Time = 2 s
• Initially at rest , that means initial velocity ( u ) = 0 m/s

TO FIND:

1. a) Acceleration of block
2. b) Force on block

SOLUTION a):-

Using the kinematic equation

s = ut + ½ at²

where

• s ➟ Distance = 8m
• u ➟ initial velocity = 0m/s
• t ➟ time taken = 2 s
• a ➟ acceleration = ?

Substituting the values we have

⟹ 8 = 0(2) + 1/2 ( a )(2)²

⟹ 8 = 2a

⟹ a = 8/2

⟾ a = 4m/s²

________________________________

SOLUTION b):-

Finding force using Newton's 2nd law

⟼ F = m × a

Where

• F ➛ force = ?
• m ➛ mass = 0.5kg
• a ➛ acceleration = 4m/s²

Substituting the values we have

➩ F = 0.5 × 4

➳ F = 2N

_______________________________

.°. Hence, Force = 2N & acceleration = 4 m/s²