Question

Find the acceleration produced on a mass of 250 kg by a force of 1250 N.Find the distance travelled from rest in 10 seconds.

Answer 1

Given:-

→ Mass of the body = 250 kg

→ Force = 1250 N

→ Initial velocity = 0

→ Time taken = 10 s

To find:-

→ Acceleration produced.

→ Distance travelled in given time.

Solution:-

By Newton's 2nd law of motion, we know that :

a = F/m

Where :-

• a is the acceleration.

• F is force.

• m is mass of the body.

=> a = 1250/250

=> a = 5 m/s²

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Now, let's calculate the distance travelled by using the 2nd equation of motion :-

s = ut + 1/2at²

Where :-

• s is the distance travelled.

• u is the initial velocity.

• a is acceleration.

• t is time taken.

=> s = 0(10) + 1/2(5)(10)²

=> s = 0 + 2.5(100)

=> s = 0 + 250

=> s = 250m

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Thus :-

• Acceleration produced is 5m/s² .

• Distance travelled is 250m .

Answer 2

Explanation:

Given:-

mass of body(m)=250kg

Force (F)=1250N

Time taken (t)=10s

initial velocity =u=0

To find:-

Acceleration =a

Distance travelled =s

Solution:-

According to Newton's 2nd law of motion

$\boxed{\sf F=ma}$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow 1250=250\times a$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow 250a=1250$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow a=\dfrac {1250}{250}$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow a=5m/s^2$

$\\\therefore\sf Acceleration \:produced\:by\:the\:man\;is\:5m/s^2.$

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According 2nd equation of kinematics

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

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow s=0 (10)+\dfrac{1}{2}5\times (10)^2$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow s=\dfrac {1}{2}5\times 100$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow s=\dfrac {500}{2}$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow s=250m$

$\\\therefore\sf Distance\:travelled \;by\:the\:man\:is\:250m.$