The expression for the acceleration of the cart for the experiment you did in la

Question

The expression for the acceleration of the cart for the experiment you did in lab is a = (m_1 g - r)/(m_1 + m_2) where m_1 + m_2 is kept constant throughout. Consider the situation when friction force is zero. What is the expression for the acceleration of the cart now? (Use the following as necessary: m_1, m_2, and a = m_1 g/(m_1 + m_2) If you now double the hanging mass m_1 (so m = 2m_1), how will the value of the denominator for the expression for the acceleration change? The denominator will double in value. The denominator value will be 2m_1 + m_2. Doubling will decrease m_2 by that amount and the total mass will have a constant value as required by the experiment.? What is the expression for the acceleration now? (Use the following as necessary: m_1 m_2, and g.) a = 2m_1 g/(2m_1 + m_2) Based on your answers to the previous parts, what conclusion can you draw about the acceleration of the cart?

Explanation / Answer

as it is given that total mass m1+m2 is constant. , Increase in m1 wil make a decrease in m2, So the total mass is m1+ m2 is contstnt , but acceleration now wil be new mass (2m1) of 1 diveded by total mass

a= 2m1/(m1+m2)

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