Question

A CONSTANT FORCE ACTS ON AN OBJECT OF MASS 10KG FOR A DURATION OF 4S. IT INCREASES THE OBJECT'S VELOCITY FROM 6M/S TO 14M/S. FIND THE MAGNITUDE OF THE APPLIED FORCE. NOW, IF THE FORCE WAS APPLIED FOR DURATION OF 10S, WHAT WOULD BE THE FINAL VELOCITY OF THE OBJECT?

Answer 1

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Answer :

• Force = 20N
• final velocity = 26m/s

Step by step explanation :

GiveN :

• Mass of body = 10kg
• Time = 4 second
• Initial velocity (u)= 6m/s
• Final velocity (v) = 14m/s

To FinD :

• Magnitude of force

Formula using :

$\large{\sf{\blue{F=\: \frac{m(v - u)}{t} }}}$

Solution :

We have given formula, Now putting the values in given formula ➺

$\large{\sf{F= \frac{m(v - u)}{t} }}$

⇒$\small{\sf{\frac{10(14 - 6)}{4} }}$

⇒10×8/4

⇒$\small{\sf{10 \times 2}}$

⇒$\small{\sf{20N}}$

↦$\large{\bf{\boxed{\green{Hence,\: the\: Force =\: 20N}}}}$

Now, if this force is applied for a

duration of 10 s (t = 10 s), then the final

velocity(v) can be calculated by using formula ⇨

Using formula :

$\large{\bf{\pink{v = u + \frac{Ft}{m} }}}$

Solution :

$\large{\sf{v = u + \frac{Ft}{m} }}$

⇒$\small{\sf{v = 6 + \frac{20 \times 10}{10} }}$

⇒$\small{\sf{v = 6 + \frac{200}{10}}}$

⇒$\small{\sf{v=\frac{60 + 200}{10}}}$

⇒$\small{\sf{v=\frac{260}{10 } }}$

$\large{\bf{\boxed{\green{v=26m/s}}}}$