The purpose of this lab investigation is to observe the relationship among the net force, mass, and acceleration of an object.
If the net force increases with a constant mass,
then the acceleration would increase,
because the force would push the object to increase the velocity.
If the mass of the cart increases with a constant net force, then the acceleration would increase
because the greater inertia of the object would cause the acceleration to decrease.
ticker timer, ticker tape, cart, masking tape, one 2-m board, marker, ruler, spring scale, three 100-g masses, two 1.0-kg masses, string,
Part A: Acceleration and Net Force
1. Verify that the equipment you intend to use is functioning properly. 2. Measure the mass of the cart and record it in the observation table. 3. Set up the apparatus so that the least net force will act on the cart. Allow the motion to occur and obtain the data required to find the acceleration α1. 4. Repeat the procedure with an increased net force. For example, you can transfer one of the 100-g masses from the cart to the string hanging over the pulley. This allows the mass of the system to remain constant. Determine the data for α2. 5. Repeat the procedure with the highest net force to determine the data for α3.
Part B: Acceleration and Mass
6. Use the data for α3 as the first set of data in this part of the experiment. Call the acceleration α4. 7. Keep the net force constant at the highest value, but add a 1.0-kg mass to the cart. Perform the trial to obtain the data for α5. 8. Add another 1.0-kg mass. Repeat the step to determine the data for α6.
TrialTotal mass of system (kg)Net force
(N [fwd])Time (s)Displacement
(m [fwd])Acceleration (m/s2 [fwd])Ratio of net force to mass (N/kg) α11.11.051/600.28960.80170.91
(c)Trial α1 : 1.0N/1.1kg = 0.9090909 . ∴ The ratio of the net force to the total mass is 0.9N/kg. Trial α2 : 2.0N/1.1kg = 1.8181818 . ∴ The ratio of the net force to the total mass is 1.8N/kg. Trial α3 : 3.0N/1.1kg = 2.7272727 . ∴ The ratio of the net force to the total mass is 2.7N/kg. Trial α5 : 3.0N/2.1kg = 1.4285714 . ∴ The ratio of the net force to the total mass is 1.4N/kg. Trial α6 : 3.0N/3.1kg = 0.9677419 . ∴ The ratio of the net force to the total mass is 1.0N/kg. (d)
m = (y2-y1)/x2-x1)
The graph has a constant slope, which indicates that the net force is directly proportional to the acceleration.
The graph is not a straight line, so the relationship between acceleration and mass is not directly proportional. (f)
m = (y2-y1)/x2-x1)
The graph has a constant slope, which represents the net force. The slope of the graph is 2.8, which is very close to the value of the net force (3.0).
(g) From the observation table, it can be seen that the magnitude of the acceleration is very close to the ratio of the net force to the mass of the system in each case. Therefore, it can be concluded that the acceleration equals to the net force divided by the mass. In essence, the equation, α=Fnet/m, can be derived from the observation.
I discover that the acceleration vector is very close to Fnet/m. According to the formula I derived in question (g), the acceleration should equal to Fnet/m. However, the points are not exactly on y=x graph because of errors.
(i) The acceleration of a cart is directly...