Block on Block

Classical Mechanics Level 2

A small block B is placed on another block A of mass 5 kg 5 \text{ kg} and length 20 cm 20 \text{ cm} . All the surfaces are frictionless.

Initially, block B is near the right end of block A as shown in the figure. A constant horizontal force of 10 N 10 \text{ N} is applied to block A.

Find the approximate time elapsed in seconds, before block B separates from A.

They will never separate Cannot be determined 0.35 0.45

View wiki
Kandarp Singh by Kandarp Singh , 22, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Kandarp Singh
Aug 1, 2016

As all the surfaces are smooth, there will be no effect of Force F on Block B.

Hence the acceleration of block A would be F = m a F=ma , F = 10 N F=10 \text{ N} , m = 5 kg m= 5 \text{ kg}

⟹ 10 = 5 × a ⟹ a = 2 m s − 2 \implies 10=5 \times a \\ \implies a=2 \text{ m s}^{-2}

Block B will lose contact when the left edge of Block A passes below the right edge of B

⟹ \implies Using 2nd equation of motion

s = 0.2 m a = 2 m s − 2 s = 1 2 a t 2 s=0.2 \text{ m}\\ a=2 \text{ m s}^{ -2 } \\ s=\frac { 1 }{ 2 } a{ t }^{ 2 } (initial velocity=0)

⟹ 0.2 = 1 2 × 2 × t 2 t 2 = 0.2 t = 0.2 ≈ 0.447 ≈ 0.45 \implies 0.2=\frac { 1 }{ 2 } \times 2 \times { t }^{ 2 }\\ { t }^{ 2 }=0.2\\ t=\sqrt { 0.2 } \approx 0.447 \approx \boxed { 0.45 }

3 Helpful 0 Interesting 0 Brilliant 0 Confused

The acceleration of the lower block due to the applied force is 2 m s − 2 2 ~m s^{-2} . How can you apply the second equation of motion to the upper block in this case? From the frame of reference of Earth(suppose inertial), the acceleration of lower block is 2 m s − 2 2 ~m s^{-2} but that of the upper one is 0 0 as no external force is applied on it (first law of motion). The correct approach was to solve this problem from the frame of lower block and since it is non-inertial, apply pseudo-force to the upper one. In this case, we obtain the acceleration of upper block as 10 / 2 = 5 m s − 2 {10}/{2} = 5 ~m s^{-2} . Now, since the separation between two points is frame independent, we can now apply the second equation of motion as:

s = u t + 1 2 a t 2 ⟹ 0.2 = ( 0 ) t + 1 2 × 5 × t 2 ⟹ t 2 = 0.08 ⟹ t ≈ 0.28 sec \begin{aligned} s & = & ut + \dfrac {1}{2} a t^2 \\ \implies 0.2 & = & (0)t + \dfrac{1}{2} \times 5 \times t^2 \\ \implies t^2 & = & 0.08 \\ \implies t & \approx & \boxed{0.28 ~ \text{sec}} \end{aligned}

Tapas Mazumdar - 4 years, 8 months ago

I have applied 2nd equation for the left edge of lower block

Kandarp Singh - 4 years, 8 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...