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

hope it helps uh :)

Answer 2

Answer:

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 (a) :-

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 :

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

sᴜʙsᴛɪᴛᴜᴛɪɴɢ ᴠᴀʟᴜᴇs

$8 = 0 \times 2 + \frac{1}{2} \times a \times 2 {}^{2} \\ 8 = 0 + \frac{1}{2} \times a \times 4 \\ 8 = 2a \\ a = \frac{8}{2} \\ a = 4ms {}^{ - 2}$

SOLUTION (b) :-

Finding force using newton's 2nd law

→F = m × a

Where..

F -- force = ?

m -- mass = 0.5kg

a -- acceleration = m/s

Substitute the values we have

→ F = 0.5×4

→ F = 2 N

_________________________________

Force = 2n

Acceleration = 4m/s²

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