A long string is stretched by 2 cm and the potential energy is V. If the spring is stretched by 10 cm, its potential energy will be
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Solution
\(v=\frac{1}{2}k(x)^{2}=\frac{1}{2}k(2)^{2}\) or \(k=z\frac{2v}{4}=\frac{v}{2}\)
\(v'=\frac{1}{2}k(10)^{2}=\frac{1}{2}\times \left ( \frac{1}{2} \right )(10)^{2}=25v\)
The engine of a truck moving along a straight road delivers constant power. The distance travelled by the truck in timet is proportional to
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Solution
\(\frac{mv^{2}}{t}=\frac{ms^{2}}{t^{3}}\)
since both P and m are constants
∴\(\frac{s^{3}}{t^{3}}\)= constant
If the force acting on a body is inversely proportional to its velocity, then the kinetic energy acquired by the body in time t is proportional to
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Solution
\(F\alpha \frac{1}{v}(given)\)
Then W=Ek=F.s
Become =Ekα\(\frac{s}{v}\Rightarrow E_{k}\alpha \frac{s}{s/t}\)
⇒Ekαt
The work donein stretching a spring of force constant k from length \(\iota _{1}\) and \(\iota _{1}\) is
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Solution
\(w=\frac{1}{2}k\iota _{2}^{2}-\frac{1}{2}k\iota _{1}^{2}=\frac{1}{2}k\left ( \iota _{2}^{2}-\iota _{1}^{2} \right )\)
Which of the following must be known in order to determine the power output of an automobile?
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Solution
Power is defined as the rate of doing work.For the automobile, the power output is the amount of work done (overcoming friction)divided by the length of time in which the work was done.
A force \(\dot{F}=\left ( 5\hat{i}+3\hat{j}+3\hat{k} \right )\) is applied over a particle which displaces it from its origin to the point \(\dot{r}=\left ( 2\hat{i}-\hat{j} \right )\)m.The work done on the particle in joule is
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Solution
\(W=\dot{F}.\dot{x}=(5\hat{i}+3\hat{j}+2\hat{k}).(2\hat{i}-\hat{j})=10-3=7\) joules
A spring of force constant 800 N/m has an extension of 5cm. The work done in extending it from 5 cm to 15 cm is
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Solution
Workdone, W=\(\frac{1}{2}k(k_{2}^{2})x_{1}^{2}\)
\(=\frac{1}{2}k\left [ (0.15)^{2}-(0.05)^{2} \right ]\)
\(=\frac{1}{2}\times 800\times 0.02=8j\)
When a U238 nucleus, originally at rest, decays by emitting an α-particle, say with speed of v m/sec, the recoil speed of the residual nucleus is (in m/sec.)
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Solution
From m1v1+m2v2=0.
\(v_{2}=\frac{m_{1}v_{1}}{m_{2}}=-\frac{4}{234}\)
A motor of 100 H.P. moves a load with a uniform speed of 72 km/hr. The forward thrust applied by the engine on the caris
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Solution
Forward thrust, F=\(\frac{p}{v}=\frac{100\times 746}{20}=3730\: N.\)
A particle moving in the xy plane undergoes a displacement of \(\dot{s}=(2\hat{i}+3\hat{j})\) while a constant force \(\dot{F}=(5\hat{i}+2\hat{j})\) N acts on the particle. The work done by the force F is
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Solution
\(W=\dot{F}.\dot{s}=(5\hat{i}+2\hat{j}).(2\hat{i}+3\hat{j})= 10 + 6 = 16 J.\)