Question

A 5 Kg cart is pushed with a force of 40 N for 7 seconds. What is its change in velocity?

Answer 1

Answer:

cart = 5kg

force =40N

time = 7 s

change of velocity u-v ➗ t

= 5kg X 40N ➗ 7s

= 200 ➗ 7

= 28.571.....

Answer 2

Given :

• Mass of the cart = 5 kg
• Force applied = 40 N
• Time taken = 7 s

To find :

• Change in velocity

Solution :

Force is given by the formula  ,

$\leadsto F = \dfrac{\Delta P }{t}$

$\leadsto F = \dfrac{ P_f - P_i }{t}$

here ,

• F = force
• Pi = initial momentum
• Pf = final momentum

we know that ,

P = mv

here ,

• P = momentum
• m = mass
• v = velocity

Hence,

$\leadsto F = \dfrac{ mv - mu }{t}$

$\leadsto F =m \dfrac{ v - u }{t}$

$\leadsto F =m \dfrac{ \Delta v }{t}$

On rearranging ,

$\leadsto \Delta v = \dfrac{ F \times t}{m}$

Now let's substitute the given values in the above equation ,

$\leadsto \Delta v = \dfrac{ 40 \times 7}{5}$

$\leadsto\boxed{ \Delta v = 56 \: m /s}$

The change in velocity is 56 m/s .