Question

A mass of 50 kg was moving with a velocity 400 m/s. A force of 40000 N is applied on the mass and its velocity is reduced to 50 m/s after some time. What is the distance travelled by the mass during this period?​

Answer 1

Answer:

The distance travelled by the mass during this period is 98.5 m.

Explanation:

Given:

• mass (m) = 50 kg
• Initial velocity (u) = 400 m/s.
• Final velocity (v) = 50 m/s.
• Force applied (F) = 40000 N

To find:

Distance travelled by the mass.

Solution:

In order to find the distance travelled, we need to find time first.

We know that,

→ a = F/m

→ a = 40000/50

→ a = 800 m/s square.

Now we know that,

Time (t) = change in velocity/time.

t = 350/800 s.

Now using the formula,

s = ut – 1/2at^2

we will find the distance travelled.

s = 400 (350/800) – (1/2) 800.

= 98.5 m

Answer 2

Given :

• Mass = 50 kg
• Initial velocity = 400 m/s
• Final velocity = 50 m/s
• Force = - 40,000 N ( retarding)

To find :

• Distance traveled

How to solve :

• First we need to find the acceleration of the mass by using Newton's second law of motion .
• As the mass is moving with uniform acceleration , we can use the third equation of motion in order to find the unknown variable (distance).

Solution :

By applying Newton's second law ,

⇒ F = ma

On rearranging ,

⇒ a = F / m

⇒ a = - 40,000 / 50

⇒ a = - 8,00 m/s ²

As per the formula ,

⇒ v² = u² + 2as

On rearranging ,

⇒ s = v² - u² / 2a

⇒ s = 2500 -  1,60,000  / 2 x- 800

⇒ s = - 1,57,500 / - 1,600

⇒ s = 98.4 m

The distance traveled by the mass during this time period is 98.4 m .