Question

purpose: in this lab activity, you will explore newtons second law. directions: at each table there will be a blue cart, a wheel on the edge of the table, and two counterweights. follow the directions in the sections below. 1) gravity of the counterweights will be used to apply a force to the cart. manipulate and/or erase the arrows to the right to draw the fbd for all the forces acting on the counterweight as the counterweight is falling toward the ground. (assume the wheels on the cart make the friction low enough to ignore.) 2) use the spring scale to measure the magnitude of the weight (fg) for each of the counterweights. this will be the force that will be applied to the cart. (assume friction is low enough to ignore.) 3) manipulate and/or erase the arrows to the right to draw the fbd for all the forces acting on the cart on the table as the counterweight is falling toward the ground. (assume friction is low enough to ignore.) 4) the cart plus the metal cylinder in the cart will have a combined mass of 0.600 kg. time how long it takes the cart to travel from the time the counterweight is released to the time the counterweight hits the ground

Explanation:

Step1: Draw FBD for counter - weight

The only two forces acting on the counter - weight as it falls are the force of gravity ($F_g$) acting downwards and the tension force ($F_T$) acting upwards. So, draw a downward - pointing arrow for $F_g$ and an upward - pointing arrow for $F_T$.

Step2: Measure weight of counter - weights

The weight of an object is given by $F_g=mg$, where $g = 9.8m/s^2$. For a 50g (or 0.05kg) counter - weight, $F_g=0.05\times9.8 = 0.49N$; for a 100g (or 0.1kg) counter - weight, $F_g=0.1\times9.8=0.98N$.

Step3: Draw FBD for cart

The forces acting on the cart are the normal force ($F_N$) acting upwards, the force of gravity ($F_{g_{cart}}$) acting downwards (but since the cart is on a horizontal surface, $F_N = F_{g_{cart}}$), and the tension force ($F_T$) acting horizontally which is equal to the weight of the counter - weight (assuming massless string and frictionless pulley). Draw an upward arrow for $F_N$, a downward arrow for $F_{g_{cart}}$, and a horizontal arrow for $F_T$.

Step4: Calculate time for cart's motion

We can use the kinematic equation $x = v_0t+\frac{1}{2}at^2$. Since the cart starts from rest ($v_0 = 0$), $x=\frac{1}{2}at^2$. First, we find the acceleration of the system using Newton's second law $F = ma$. The force acting on the system is the weight of the counter - weight, and the total mass of the system is the mass of the cart plus the metal cylinder ($m_{total}=0.600kg$). Once we know the acceleration $a=\frac{F_{counter - weight}}{m_{total}}$, and if we know the distance $x$ the cart travels, we can solve for $t=\sqrt{\frac{2x}{a}}$.

Answer:

1. For the counter - weight FBD: Draw a downward arrow labeled $F_g$ and an upward arrow labeled $F_T$.
2. Weight of 50g counter - weight is 0.49N, weight of 100g counter - weight is 0.98N.
3. For the cart FBD: Draw an upward arrow labeled $F_N$, a downward arrow labeled $F_{g_{cart}}$, and a horizontal arrow labeled $F_T$.
4. Use $t=\sqrt{\frac{2x}{a}}$ where $a=\frac{F_{counter - weight}}{m_{total}}$ and $m_{total}=0.600kg$ to find the time, after measuring the distance $x$ the cart travels.