Question

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

Answer 1

Given :

Initial velocity = zero

Mass of block = 500g

Applied force = F

Distance travelled in 2s = 8 m

To Find :

• Acceleration
• Value of applied force

Solution :

❖ Assuming that body has constant acceleration throughout the motion$.$

Acceleration of block can be easily calculated by applying equation of kinematics.

Second equation of kinematics :

• d = ut + 1/2 at²

» d denotes distance

» u denotes initial velocity

» t denotes time

» a denotes acceleration

By substituting the given values;

➙ d = ut + 1/2 at²

➙ 8 = (0 × 2) + 1/2 a (2)²

➙ 8 = 0 + 4a/2

➙ 8 = 2a

➙ a = 8/2

➙ a = 4 m/s²

♦ As per newton's second law of motion, force is measured as the product of mass and acceleration.

Mathematically, F = m a

[Mass (m) = 500g = 0.5kg]

➙ F = 0.5 × 4

➙ F = 2 N

Answer 2

${\rm{\underline{\underline{Question:-}}}}$

→ A block of mass 500g is pulled from rest on a horizontal frictionless bench by a steady force F and travels 8m in 2s find

A. acceleration

B. the value of F

${\rm{\underline{\underline{Given:-}}}}$

✩ Mass of the block = 500 grams

✩ Distance travelled in 2s = 8m

✩ Initial velocity of the block = 0m/s

✩ Force applied on the block = F

${\rm{\underline{\underline{To \;Find:-}}}}$

✩  Acceleration

✩  The value of F \ force applied

${\rm{\underline{\underline{Solution:-}}}}$

→ As we know that

${\red{\boxed{\underline{Force=Mass \times Acceleration}}}}\\\\\\\\{\red{\boxed{\underline{d = ut \times \dfrac{1}{2}at^{2}}}}}}$

$\sf Where , \\\\ d = distance \\\\ u = initial \; velocity \\\\ t = time \\\\a = acceleration$

→ let's solve by substituting the values !!

$\sf \rightarrow d = ut + \dfrac{1}{2}at^{2} \\\\\\\rightarrow 8 = (0 \times 2 ) + \dfrac{1}{2} \times a(2)^{2} \\\\\\\rightarrow 0 + \dfrac{4a}{2} \\\\\\\rightarrow 8 = 2a \\\\\rightarrow a = \dfrac{8}{2} \\\\\rightarrow a = 4 \; m/s^{2}$

∴ Acceleration = 4 m/s²

→ For finding the value of force applied let's covert the mass given into kg

$\sf 1 \; g = \dfrac{1}{1000} \; kg \\\\\\500 \; g = \dfrac{500}{1000} \; kg \\\\\\\longrightarrow 0.5 kg$

→ Now , finding the value of force applied :

$\sf \rightarrow Force = Mass \times Acceleration\\\\\rightarrow Force = 0.5 \times 4 \\\\\rightarrow Force = 2 \; Newton$

∴ Force applied on the block is 2 N

___________

All Done !! :D