Question

pls answer correctly​

Answer 1

In this type of questions it is best to apply Work - Energy Theorem. It states that the work done by all the forces will be equal to the change in kinetic energy over the period of time.

$x = 3t - 4 {t}^{2} + {t}^{3}$

$= > v = \dfrac{dx}{dt}$

$= > v = 3 - 8t + 3 {t}^{2}$

At t = 0 secs , velocity be v1 :

$v1 = 3 - 0 + 0 = 3 \: m {s}^{ - 1}$

At t = 4 secs , velocity be v2 :

$v2 = 3 - 32 + 48 = 19 \: m {s}^{ - 1}$

So change in kinetic energy will be :

$\Delta KE = \dfrac{1}{2} m \{ {(v2)}^{2} - {(v1)}^{2} \}$

$\Delta KE = \dfrac{1}{2} \times \bigg(\dfrac{30}{1000} \bigg) \times \{ {(19)}^{2} - {(3)}^{2} \}$

$\Delta KE = \dfrac{1}{2} \times \bigg(\dfrac{30}{1000} \bigg) \times \{ 361 - 9 \}$

$\Delta KE = \dfrac{1}{2} \times \bigg(\dfrac{30}{1000} \bigg) \times \{ 352 \}$

$\Delta KE = \dfrac{528}{100}$

$\Delta KE = 5.28 \: joule$

So work done is equal to change in Kinetic Energy ;

So final answer :

Work done is 5.28 J

Answer 2

Answer:

• The work done (W) is 5.35 J

Given:

1. Mass of the particle (M) = 30 g = 0.03 Kg
2. Relation:- x = 3 t - 4 t² + t³
3. Time period (t) = 4 sec

Explanation:

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From the relation we know,

x = 3 t - 4 t² + t³

Differentiating the equation w.r.t time to get velocity.

$\displaystyle\Rightarrow\sf \dfrac{dx}{dt}=\dfrac{d(3t-4t^{2}+t^{3})}{dt}\\\\\\\Rightarrow\sf v=\dfrac{d(3t-4t^{2}+t^{3})}{dt}\\\\\\\Rightarrow\sf v=3t^{(1-1)}-(4\times2)t^{(2-1)}+(1\times3)t^{(3-1)}\\\\\\\Rightarrow\sf v = 3t^{0}-8t^{1}+3t^{2}\\\\\\\Rightarrow\bf\underline{v=3-8t+3t^{2}}$

Differentiating the equation w.r.t time to get acceleration.

$\displaystyle\Rightarrow\sf \dfrac{dv}{dt}=\dfrac{d(3-8t+3t^{2})}{dt}\\\\\\\Rightarrow\sf a=\dfrac{d(3-8t+3t^{2})}{dt}\\\\\\\Rightarrow\sf a=0-8t^{(1-1)}+(3\times2)t^{(2-1)}\\\\\\\Rightarrow\sf a= -8+6t\\\\\\\Rightarrow\bf\underline{a=-8+6t}$

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Before solving, we need to see a case,

As we know,

$\displaystyle\Rightarrow\sf \dfrac{dx}{dt}=\dfrac{d(3\;t-4\;t^{2}+t^{3})}{dt}\\\\\\\Rightarrow\sf dx=3\;t-4\;t^{2}+t^{3}.dt\\\\\\\Rightarrow\sf dx=3\;t-4\;t^{2}+t^{3}.dt\quad\dfrac{\quad}{}(1)$

Now, From Work done formula,

$\displaystyle \bigstar\;\underline{\boxed{\sf W=\displaystyle\int \sf F.dx}}$

Here,

• W Denotes Work done.
• F Denotes Force.
• d x Denotes small displacement.

Substituting the values,

$\displaystyle\Rightarrow\sf W=\displaystyle\int \sf F.dx\\\\\\\Rightarrow\sf W=\displaystyle\int \sf (M \;a).dx\\\\\\\Rightarrow\sf W=\displaystyle\int \sf \bigg\lgroup0.03 \times (-8+6\;t)\bigg\rgroup.dx$

Substituting the value of dx from equation (1)

$\displaystyle\Rightarrow\sf W=\displaystyle\int \sf \bigg\lgroup0.03 \times (-8+6\;t)\bigg\rgroup \bigg\lgroup 3\;t-4\;t^{2}+t^{3}.dt\bigg\rgroup\\\\\\\Rightarrow\sf W=0.03 \displaystyle\int \sf \bigg\lgroup-8+6\;t\bigg\rgroup \bigg\lgroup 3\;t-4\;t^{2}+t^{3}.dt\bigg\rgroup$

Multiplying both gives,

$\displaystyle\Rightarrow\sf W=0.03 \displaystyle\int \sf \bigg\lgroup18t^{3}-72t^{2}+82t-24\bigg\rgroup .dt$

Applying the limits,

• i.e. 0 to 4

$\displaystyle\Rightarrow\sf W=0.03 \displaystyle\int^{4}_{0} \sf \bigg\lgroup18t^{3}-72t^{2}+82t-24\bigg\rgroup .dt\\\\\\\Rightarrow\sf W=0.03 \times \Bigg\lgroup18\times\dfrac{t^{4}}{4}-72\times\dfrac{t^{3}}{3}+82\times\dfrac{t^{2}}{2}-24\times t^{1}\Bigg\rgroup^{4}_{0}\\\\\\\Rightarrow\sf W=0.03\times \Bigg\lgroup18\times\dfrac{(4)^{4}}{4}-72\times\dfrac{(4)^{3}}{3}+82\times\dfrac{4^{2}}{2}-24\times 4\Bigg\rgroup$

$\displaystyle\Rightarrow\sf W=0.03\times \Bigg\lgroup18\times\dfrac{256}{4}-72\times\dfrac{64}{3}+82\times\dfrac{16}{2}-24\times 4\Bigg\rgroup\\\\\\\Rightarrow\sf W=0.03\times \Bigg\lgroup18\times64-72\times21.3+82\times8-24\times 4\Bigg\rgroup\\\\\\\Rightarrow\sf W=0.03\times \Bigg\lgroup1152-1533.6+656-96\Bigg\rgroup$

$\displaystyle\Rightarrow\sf W=0.03\times 178.4\\\\\\\Rightarrow\sf W=5.35\\\\\\\Rightarrow \large{\underline{\boxed{\red{\sf W=5.35\;J}}}}$

∴ The work done (W) is 5.35 J

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