Physics MCQs for NEET — Practice Questions with Answers

The tension in the supporting cable is the force required to provide the necessary acceleration to the combined mass of the lift and the person. Using Newton's second law, F = ma = (60 + 940) kg × (10 + 1) m/s^2 = 11000 N.

The tension in the supporting cable is the force required to accelerate the lift upwards. Using Newton's second law, F = ma, where F is the tension (28000 N), m is the mass of the lift (2000 kg), and a is the acceleration. Therefore, a = 28000 N / 2000 kg = 14 m/s^2 upwards.

According to Newton's second law of motion, force = mass $\times$ acceleration.

Here,  $\overrightarrow{F}=6\hat{i}-8\hat{j}+10\hat{k}$

         $\left|F\right|=\sqrt{36+64+100}$

             = $10\sqrt{2}N$

        $a=1 m{s}^{-2}$

$\therefore m=\frac{10\sqrt{2}}{1}=10\sqrt{2}kg$

Balancing the forces, we get

        $Mg-N=M\frac{v^2}{R}$

For weightlessness, N = 0

$\therefore \frac{M{v}^2}{R}=Mg$

where R is the radius of curvature and v is the speed of car.

Therefore,                $v=\sqrt{Rg}$

Putting the values, R=20m, g=10.0m/$s^2$

So,  $v=\sqrt{20\times 10.0}=14.14m/s^2$

Thus, the speed of the car at the top of the hill is between 14m/s and 15m/s.

Note: The roller coaster is a popular amusement ride developed for amusement parks and modern theme parks.

To keep the belt moving at the same speed, the force required will be equal to change in its momentum.

Force required, F=$\frac{d\left(mv\right)}{dt}$

                        =v$\left(\frac{dm}{dt}\right)=Mv$

As velocity v is constant, hence,

                       F = Mv newton

Here 

$$\begin{gathered} k=\frac{F}{x}=\frac{10}{1\times {10}^{-3}}={10}^4 N/m \\ W=\frac{1}{2}k{x}^2=\frac{1}{2}\times {10}^4\times {\left(40\times {10}^{-3}\right)}^2=8J \end{gathered}$$

Since force is same

$$\begin{gathered} \mathrm{F}= {\mathrm{k}}_1{\mathrm{x}}_1 = {\mathrm{k}}_2{\mathrm{x}}_2 \\ \mathrm{So}, \frac{{\mathrm{x}}_1}{{\mathrm{x}}_2}=\frac{ {\mathrm{k}}_2}{{\mathrm{k}}_1} \\ \mathrm{Also}, \frac{{\mathrm{U}}_1}{{\mathrm{U}}_2}=\frac{{\displaystyle \frac{1}{2}}{\mathrm{k}}_1{{\mathrm{x}}^2}_1}{\frac{1}{2}{\mathrm{k}}_2{{\mathrm{x}}^2}_2} =\frac{{\mathrm{k}}_2}{{\mathrm{k}}_1}=\frac{3000}{1500}=\frac{2}{1} \end{gathered}$$

Using work-energy theorem

$$\begin{gathered} W_{All force}={\left(KE\right)}_f-{\left(KE\right)}_i \\ =\frac{1}{2}\times 10\times {\left(30\right)}^2-\frac{1}{2}\times 10\times {\left(20\right)}^2 \\ =5\left[50\times 10\right]=2500 J \end{gathered}$$

$$\begin{gathered} P= F.v \\ P=ma.v \\ P=mv\frac{dv}{dt} \\ Pdt=mvdv \\ P\int dt=m\int vdv \\ P.t=m\frac{v^2}{2} \\ v=\sqrt{\frac{2Pt}{m}} \\ dx=\sqrt{\frac{2Pt}{m}}dt \\ \int dx=\sqrt{\frac{2P}{m}}\int \sqrt{t}dt \\ x=\sqrt{\frac{2P}{m}}\times \frac{2}{3}{t}^{3/2} \end{gathered}$$