Concepts Of Physics MCQ Edition [Volume 1]PhysicsNewton's Laws of Motion
As shown in the figure, an elevator is descending with an acceleration of 2 m/s ^2 . Block A has a mass of 0.5 kg . Determine the force exerted by block A on block B .
Options
- A6 N
- B1 N
- C4 N
- D5 N
Correct answer
C. 4 N
Step-by-step solution
Let the normal force between block A and block B be N . Considering the free body diagram of block A, the forces acting on it are its weight ( m_A g ) downwards and the normal reaction ( N ) upwards. Since the elevator is accelerating downwards with an acceleration a = 2 m/s ^2 , the equation of motion for block A is: m_A g - N = m_A a N = m_A(g - a) Given m_A = 0.5 kg , a = 2 m/s ^2 , and taking g = 10 m/s ^2 : N = 0.5 (10 - 2) N = 0.5 8 = 4 N By Newton's third law, the force exerted by block A on block B is equal