Calculating Stopping Time for a Body under Constant Retarding Force

Question:

A constant retarding force of 50 N is applied to a body of mass 20 Kg moving initially with a speed of 15 ms-1 .How long does the body take to stop?

Solution: Stopping time calculation:

Calculating the acceleration:

$${Acceleration =\frac{Force}{Mass}}$$

$${Acceleration =\frac {50 \ N}{20 \ Kg}}$$

$${Acceleration= 2.5 \ \frac{m}{s^2}}$$

Retardation:

Retardation is -2.5 ms-2

Since the force is retarding, the acceleration is negative.

Time taken to stop:

$${v=u+at}$$

Initial velocity u =15 ms-1

Final velocity v = 0

Acceleration a = -2.5 ms-2 ( it is retarding}

Time taken =?

$${t=\frac{v-u}{a}}$$

$${t=\frac{0-15}{-2.5}}$$

$${t=6 sec}$$

Conclusion:

The body takes 6 seconds to stop under the action of a constant retarding force of 50 N.


En savoir plus sur eduPhysics

Subscribe to get the latest posts sent to your email.

🚀 Install PhysicsAce' App
Install PhysicsAce' App
WhatsApp Chat With Us
Return & Refund Policy | All purchases are final. Replacements for defective or incorrect items only.
Return and Refund Policy

En savoir plus sur eduPhysics

Abonnez-vous pour poursuivre la lecture et avoir accès à l’ensemble des archives.

Poursuivre la lecture

En savoir plus sur eduPhysics

Abonnez-vous pour poursuivre la lecture et avoir accès à l’ensemble des archives.

Poursuivre la lecture