Students work together during an experiment about Newton’s laws. The students use a setup that consists of a cart of known mass
connected to one end of a string that is looped over a pulley of negligible friction, with its other end connected to a hanging mass. The cart is initially at rest on a horizontal surface. Students have access to common laboratory equipment to make measurements of components of the system. The students double the mass that hangs from the string. They also replace the original cart with a new cart that has double the mass. By doubling both masses, how will the tension in the string and the acceleration of the cart change?
1 answer:
Answer:
The tension in the string will increase by a factor of two, while the acceleration will not change.
Explanation:
To approach this problem, we begin by sketching the situation and creating free body diagrams for both masses involved. (Refer to the attached image)
Let’s first examine the free body diagram for the cart. Assuming there’s no friction between the cart and the horizontal surface, and focusing solely on horizontal forces, we can write the force balance as follows:
Given that tension is the only horizontal force on the cart, we state:
Next, consider the hanging mass, taking downward as the positive direction. Summing forces yields:
The two forces acting here are gravity and tension, so:
Combining the two equations results in:
Solving for acceleration gives:
This represents the acceleration for the original setup. What happens if both masses are doubled? Let’s analyze:
Factoring 2 out of the denominator produces:
Which simplifies to:
You can observe this matches the original acceleration, confirming it remains unchanged.
Regarding tension, by substituting the doubled masses into the initial force expression:
The initial tension was:
Therefore, doubling the masses causes the tension to double as well.
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Here is an image displaying the correct answers.
Answer:
Jari
Explanation:
To determine who is traveling faster, we need to evaluate their gradients. A steeper slope indicates a higher speed.
For Jari's path, starting point is (0, 0) and (6, 7) is another point.
The gradient is the difference in y divided by the difference in x:
Change in y=7-0=7
Change in x=6-0=6
Thus, the slope equals 7/6.
For Jade, her first point is (0, 10) and another is (6, 16).
Change in y=16-10=6
Change in x=6-0=6
Thus, the slope equals 6/6=1.
It's evident that 7/6 exceeds 6/6 or 1, proving Jari is quicker than Jade.
