Question

10.A constant force acts on an object of mass 5 kg for duration of 2 second. It increases the
object's velocity from 3 m/s to 7m/s. Find the magnitude of the applied force. Now if the
force were applied for a duration of 5 seconds, what would be the final velocity of the
object?​

Answer 1

Answer:

F=10N,v=13m/s

Explanation:

$f = ma \\ v = u + at \\ v = 7 \: \: \: \: u = 3 \: \: \: t = 2sec \\ 7 = 3 + a2 \\ 4 = 2a \\ a = 2 \frac{m}{ {s}^{2} } \\ f = 5 \times 2 = 10 \ \\ now \: for \: t = 5 \\ v = 3 + 2 \times 5 \\ v = 13mper s$

Answer 2

Given :

• A constant force acts on an object of mass 5 kg for duration of 2 second.
• The object's velocity from 3 m/s to 7m/s. Find the magnitude of the applied force.
• The force were applied for a duration of 5 seconds

To find :

• The final velocity of the object ?

Solution :

$\sf Mass = 5 kg$

$\sf t¹ = 2s$

$\sf Intital \: velocity \: u = 3m/s$

$\sf Final \: velocity \: v = 7m/s$

t² = 5s

So,

Let the source will be F

Let , the acceleration be a

So ,

$\sf a = \frac{(v - u)}{t} = \frac{( 7 - 3}{2} = 2m \: per \: second$

So , the magnitude of the applied force is 10N

And ,

The final velocity after 5s is v

So,

v = u + at

v = 3 + 2 × 5

v = 13m/s

The final velocity will be after 5s 13m/s