You visited us 1 times! Enjoying our articles? Unlock Full Access!

Efficiency of a Machine

Two springs P...

Question

Two springs $P$ and $Q$ of force constants $k_{P}$ and $k_{Q} (k_{Q} = \frac{k_{P}}{2})$ are stretched by applying forces of equal magnitude. If the energy stored in $Q$ is $E$ , then the energy stored in $P$ is

E

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{E}{8}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{E}{4}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{E}{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $\frac{E}{2}$
$\begin{matrix} U & = \frac{1}{2} k x^{2} U_{p} & = \frac{1}{2} k_{p} x_{p}^{2} U_{Q} & = \frac{1}{2} k_{Q} x_{Q}^{2} \end{matrix}$

$F_{p} = F_{q}$

So, $\frac{U_{p}}{U_{Q}} = \frac{x_{p}}{x_{Q}}$

$∵ F_{p} = - k_{p} x_{p}$
and $F_{Q} = - k_{Q} x_{Q}$

$⟹ \frac{k_{p}}{k_{Q}} = \frac{x_{Q}}{x_{p}} = 2$

$⟹ U_{p} = \frac{E}{2}$

Suggest Corrections

Similar questions

Q. Two springs

P

and

Q

of force constants

k_{P}

and

k_{Q} (k_{Q} = \frac{k_{P}}{2})

are stretched by applying forces of equal magnitude. If the energy stored in

Q

E

, then the energy stored in

P

Two springs A and B ( $k_{A}$ = 2 $k_{B}$ ) are stretched by applying forces of equal magnitudes at the four ends. If the energy stored in A is E, that in B is

Q. Two springs A and B(k_A = 2k_B) are stretched by applying forces of equal magnitudes a the four ends. If the energy stored in A is E, that in B is_A
(a) E/2
(b) 2E
(c) E
(d) E/4

Q. Two springs A and B having spring constants

k_{A}

and

k_{B} (k_{A} = 4 k_{B})

are stretched by applying force of equal magnitude. If energy stored in spring A is

E_{A}

, then energy stored in B will be:

Q. Two springs

A

and

B

are stretched by applying forces of equal magnitudes. If the energy stored in

A

is E, then the energy stored in B will be-
(Given relation between spring constants of

A

and

B

(k_{A} = 2 k_{B})

)

Join BYJU'S Learning Program

Two springs P and Q of force constants kP and kQ(kQ=kP2) are stretched by applying forces of equal magnitude. If the energy stored in Q is E, then the energy stored in P is

The correct option is D E2U=12kx2Up=12kpx2pUQ=12kQx2Q Fp=Fq So, UpUQ=xpxQ ∵Fp=−kpxp and FQ=−kQxQ ⟹kpkQ=xQxp=2 ⟹Up=E2

Two springs $P$ and $Q$ of force constants $k_{P}$ and $k_{Q} (k_{Q} = \frac{k_{P}}{2})$ are stretched by applying forces of equal magnitude. If the energy stored in $Q$ is $E$ , then the energy stored in $P$ is