SATHEE: Work Power and Energy 1 Question 2

Work Power and Energy 1 Question 2

3. A force acts on a $2 k g$ object, so that its position is given as a function of time as $x = 3 t^{2} + 5$ . What is the work done by this force in first 5 seconds?

(a) $850 J$

(b) $900 J$

(d) $875 J$

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Answer:

Correct Answer: 3. (b)

Solution:

Here, the displacement of an object is given by

$x = (3 t^{2} + 5) m$

Therefore, velocity $(v) = \frac{d x}{d t} = \frac{d (3 t^{2} + 5)}{d t}$

$v = 6 t m / s$

The work done in moving the object from $t = 0$ to $t = 5 s$

$W = \int_{x_{0}}^{x_{5}} F \cdot d x$

The force acting on this object is given by

$\begin{aligned} F & = m a = m \times \frac{d v}{d t} \\ = m \times \frac{d (6 t)}{d t} [∴ using (i)] \\ F & = m \times 6 = 6 m = 12 N \end{aligned}$

Also, $x_{0} = 3 t^{2} + 5 = 3 \times (0)^{2} + 5 = 5 m$

and at $t = 5 s$ ,

$x_{5} = 3 \times (5)^{2} + 5 = 80 m$

Put the values in Eq. (ii),

$W = 12 \times \int_{x_{0}}^{x_{5}} d x = 12 [80 - 5]$

$W = 12 \times 75 = 900 J$

Alternative Method

To using work - kinetic energy theorem is,

$\begin{aligned} W & = Δ K \cdot E = \frac{1}{2} m (v_{f}^{2} - v_{i}^{2}) \\ = \frac{1}{2} m \times (30^{2} - 0^{2}) = \frac{1}{2} \times 2 \times 900 = 900 J \end{aligned}$

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अगला

Work Power and Energy 1 Question 2

3. A force acts on a $2 k g$ object, so that its position is given as a function of time as $x = 3 t^{2} + 5$ . What is the work done by this force in first 5 seconds?

Answer:

Alternative Method

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

Work Power and Energy 1 Question 2

3. A force acts on a 2kg 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?

Answer:

Alternative Method

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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3. A force acts on a $2 k g$ object, so that its position is given as a function of time as $x = 3 t^{2} + 5$ . What is the work done by this force in first 5 seconds?