Bob pushes on a frictionless cart of mass 22.5 kg and the cart has an acceleration of 1.3 m/s2. If Bob now pushes twice as hard on a cart with twice the mass, what will be the acceleration of the heavier cart (in m/s²)? Your Answer: Answer
Gravitational force
In nature, every object is attracted by every other object. This phenomenon is called gravity. The force associated with gravity is called gravitational force. The gravitational force is the weakest force that exists in nature. The gravitational force is always attractive.
Acceleration Due to Gravity
In fundamental physics, gravity or gravitational force is the universal attractive force acting between all the matters that exist or exhibit. It is the weakest known force. Therefore no internal changes in an object occurs due to this force. On the other hand, it has control over the trajectories of bodies in the solar system and in the universe due to its vast scope and universal action. The free fall of objects on Earth and the motions of celestial bodies, according to Newton, are both determined by the same force. It was Newton who put forward that the moon is held by a strong attractive force exerted by the Earth which makes it revolve in a straight line. He was sure that this force is similar to the downward force which Earth exerts on all the objects on it.
![**Problem Statement:**
Bob pushes on a frictionless cart of mass 22.5 kg and the cart has an acceleration of 1.3 m/s². If Bob now pushes twice as hard on a cart with twice the mass, what will be the acceleration of the heavier cart (in m/s²)?
**Solution:**
To solve this problem, apply Newton’s second law, \( F = ma \).
1. **Original Scenario:**
- Mass of cart, \( m_1 = 22.5 \) kg
- Acceleration, \( a_1 = 1.3 \) m/s²
- Force applied by Bob, \( F = m_1 \times a_1 \)
2. **New Scenario:**
- New mass of cart, \( m_2 = 2 \times 22.5 = 45 \) kg
- New force applied by Bob, \( 2F = 2 \times m_1 \times a_1 \)
- New acceleration, \( a_2 \)
The force in the second scenario is \( 2F \), and using the new mass \( m_2 \):
\[ 2F = m_2 \times a_2 \]
Substituting in the expressions for \( F \) and \( m_2 \):
\[ 2(m_1 \times a_1) = 45 \times a_2 \]
Solving for \( a_2 \):
\[ a_2 = \frac{2 \times 22.5 \times 1.3}{45} \]
\[ a_2 = \frac{58.5}{45} \]
\[ a_2 \approx 1.3 \text{ m/s²} \]
Thus, the acceleration of the heavier cart is approximately 1.3 m/s². The acceleration remains the same since both the force and the mass have doubled, maintaining the same ratio \( F/m \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa8f55c18-2057-4361-9aa0-e509a1f08c5c%2F2a9f1983-20e7-4697-8c39-872917b10153%2Fdtwcrp4_processed.jpeg&w=3840&q=75)
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