Question

8.
1) uw
A force acts on a 2 kg object so that its
position is given as a function of time as x =
3t² + 5. What is the work done by this force
in first 5 seconds?
1) 850J 2) 900J 3) 950J4) 875J​

Answer 1

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From the Question,

• Mass of the Object,m = 2 Kg

• Time,t = 5s

Distance of the object is given by:

$\sf{x = 3t {}^{2} \: + \: 5 }$

Differentiating x w.r.t t,we get velocity of the object:

$\sf{v = \frac{dx}{dt} } \\ \\ \rightarrow \: \sf{v = \frac{d(3t {}^{2} + 5)}{dt} } \\ \\ \rightarrow \: \huge{\sf{v = 6t}}$

Differentiating v w.r.t to t,we get:

$\sf{a = \frac{dv}{dt} } \\ \\ \rightarrow \: \sf{a = \frac{d(6t)}{dt}} \\ \\ \rightarrow \: \huge{ \sf{a = 6ms {}^{ - 2} }}$

Putting t = 5s,we obtain:

$\sf{v = 30 m/s}$

From Work Energy Theorem,

$\huge{\boxed{\sf{W = \frac{1}{2}m{v}^{2}}}}$

Putting the values,we get:

W = ½(2)(30)²

→ W = 900 J

Thus,the work done to move the object is 900J