Question

Assume the three blocks (m. = 1.0 kg, m = 20 kg and m = 40 ko) portrayed in the figure below move on a frictionless surface and a force F: 36w acts as shown on the 4.0 kg block.
a) Determine the acceleration given this system (in m/s2 to the right). m/s2 (to the right)
b) Determine the tension in the cord connecting the 4.0 kg and the 1.0 kg blocks in N). Determine the force exerted by the 1.0 kg block on the 2.0 kg block (in N). N (a) What If How would your answers to parts (a) and (b) of this problem change if the 2.0 kg block was now stacked on top of the 1.0 kg block? Assume that the 2.0 kg block sticks to and does not slide on the 1.0 kg block when the system is accelerated.
(Enter the acceleration in m/s2 to the right and the tension in N.) acceleration m/s (to the right) tension

Answer 1

Answer:

a) 5.143 m/s^2

b) T = 15.43 N

c) Fr = 10.29 N

d) 5.143 m/s^2 , T = 15.43 N

Explanation:

Given:-

- The mass of left most block, m1 = 1.0 kg

- The mass of center block, m2 = 2.0 kg

- The mass of right most block, m3 = 4.0 kg

- A force that acts on the right most block, F = 36 N

Solution:-

a)

- For the first part we will consider the three blocks with masses ( m1 , m2 , and m3 ) as one system on which a force of F = 36 N is acted upon. The masses m1 and m3 are connected with a string with tension ( T ) and the m1 and m2 are in contact.

- We apply the Newton's second law of motion to the system with acceleration ( a ) and the combined mass ( M ) of the three blocks as follows:

                       $F = M*a\\\\36 = ( 1 + 2 + 4 )*a\\\\a = \frac{36}{7}\\\\a = 5.143 \frac{m}{s^2}$

Answer: The system moves in the direction of external force ( F ) i.e to the right with an acceleration of 5.143 m/s^2

b)

- The blocks with mass ( m1 and m3 ) are connected with a string with tension ( T ) with a combined acceleration of ( a ).

- We will isolate the massive block ( m3 ) and notice that two opposing forces ( F and T ) act on the block.

- We will again apply the Newton's 2nd law of motion for the block m3 as follows:

                       $F_n_e_t = m_3 * a\\\\F - T = m_3 * a\\\\36 - T = 4*5.143\\\\T = 36 - 20.5714\\\\T = 15.43 N$

Answer:- A tension of T = 15.43 Newtons acts on both blocks ( m1 and m3 )

                       

c)

- We will now isolate the left most block ( m1 ) and draw a free body diagram. This block experiences two forces that is due to tension ( T ) and a reaction force ( Fr ) exerted by block ( m2 ) onto ( m3 ).

- Again we will apply the the Newton's 2nd law of motion for the block m3 as follows:

                         $F_n_e_t = m_1*a\\\\T - F_r = m_1*a\\\\15.43 - F_r = 1*5.143\\\\F_r = 15.43 - 5.143\\\\F_r = 10.29 N$

- The reaction force ( Fr ) is contact between masses ( m1 and m2 ) exists as a pair of equal magnitude and opposite direction acting on both the masses. ( Newton's Third Law of motion )

Answer: The block m2 experiences a contact force of ( Fr = 10.29 N ) to the right.

d)

- If we were to stack the block ( m2 ) on-top of block ( m1 ) such that block ( m2 ) does not slip we the initial system would remain the same and move with the same acceleration calculated in part a) i.e 5.143 m/s^2

- We will check to see if the tension ( T ) differs or not as the two block ( m1 and m2 ) both experience the same Tension force ( T ) as a sub-system. with a combined mass of ( m1 + m2 ).

- We apply the Newton's 2nd law of motion for the block m3 as follows:

                            $T = ( m_1 + m_2 ) *a\\\\T = ( 1 + 2 ) * 5.143\\\\T = 15.43 N$

Answer: The acceleration of the whole system remains the same at a = 5.143 m/s^2 and the tension T = 15.43 N also remains the same.