Question

A force of magnitude F is exerted on the leftmost face of two blocks sitting next to each other on a slippery surface, with two of their faces touching. The mass of the block on the right is m, and the mass of the block on the left is 2m. (a) What is the magnitude of the acceleration of each block? (b) What are the magnitude and direction of the contact force the left block exerts on the right block? (c) What are the magnitude and direction of the contact force the right block exerts on the left block? (d) How do your answers to parts a, b, and c change if the positions of the two blocks are interchanged

Answer 1

Answer:

Part a)

$a = \frac{F}{3m}$

Part b)

$F_n = \frac{F}{3}$

Direction = Towards Right

Part c)

$F_n' = \frac{F}{3}$

direction = Towards Left

Part d)

then acceleration must be same

so answer of Part a) will not change but answer of Part b) and Part c) will change

so new value of contact force will be

$F_n = \frac{2F}{3}$

Explanation:

Part a)

By Newton's law we have

$F = (m_1 + m_2) a$

here we have

$F = (m + 2m) a$

$a = \frac{F}{3m}$

Part b)

Force due to left block on right block is given as

$F_n = m a$

$F_n = m\times \frac{F}{3m}$

$F_n = \frac{F}{3}$

Direction = Towards Right

Part c)

By Newton's III law we know that every force has equal and opposite reaction

so we have

$F_n' = \frac{F}{3}$

direction = Towards Left

Part d)

When position of two blocks are interchanged

then acceleration must be same

so answer of Part a) will not change but answer of Part b) and Part c) will change

so new value of contact force will be

$F_n = 2m\times \frac{F}{3m}$

$F_n = \frac{2F}{3}$