Question

An airplane moves 100 m/s as it travels around a vertical circular loop which has a 1.0-km radius. What is the magnitude of the resultant force on the 50-kg pilot of this plane at the bottom of this loop? A 4.0-kg block sliding on a horizontal frictionless surface is attached to one end of a horizontal spring (k = 200 N/m) which has its other end fixed. If the block has a speed of 4.0 m/s as it passes through the equilibrium position, what is its speed when it is 20 cm from the equilibrium position?

Answer 1

7). Velocity of plane, v= 100m/s

Radius, r= 1 km

Mass of pilot, M= 50 kg

Centripetal Force acting towards centre( vertically upwards direction when plane is at bottom) on pilot is,

  

Weight of pilot acting in downward, $W=Mg= 50 \times 9.8= 490 N$

Thus Net force acting on pilot,   ( towards the centre of loop)

8). Mass of block, m= 4 kg

Spring Constant, k= 200 N/m

Velocity of block at equilibrium position, v= 4 m/s

Let block initially compressed the spring to a distance x

thus Velocity gained by block will be due to the energy provided by block. There will be complete conversion of energy in this due to frictionless surface.

Hence

  

   $x= \sqrt{mv^2/k}= \sqrt{ 4\times 4^2/200}= \sqrt{0.32}= 0.56 m$

Now force exerted by spring on block is, F_s= kx

Also by newton's laws of motion, F_b= ma

Equating both forces as force on block is due to spring only.

hence $ma=kx\Rightarrow a= kx/m$

Now at equilibrium position, initial velocity, u= 4m/s

acceleration, a= 28.28 m/s²

Distance, S= 20cm= 0.2m

let final velocity at this much distance be v

thus by using equations of motion,

$v^2-u^2=2aS$

$v^2-(4)^2=2(28.28)(0.2)$

   $v=\sqrt{11.31+16}= \sqrt{27.31}= 5.22 m/s$   ( ANS)