A dynamics cart with a mass of 1.70kg on a horizontal table is attached to a suspended mass of 0.40kg and released from rest. If a frictional of 2.0N opposes the motion of the cart how far does the cart travel in 1.5s
To determine the distance traveled by the cart in 1.5 seconds, we need to calculate the acceleration of the system and then use the equations of motion.
First, let's determine the acceleration of the system. The net force acting on the system is the sum of the force due to friction and the force due to the tension in the rope:
Net force = Force due to friction + Force due to tension
Since the cart is released from rest, the force due to tension in the rope is equal to the weight of the suspended mass:
Force due to tension = Weight of the suspended mass
Force due to tension = mass of the suspended mass × acceleration due to gravity
Force due to tension = (0.40 kg) × (9.8 m/s²)
Force due to tension ≈ 3.92 N
Now, let's calculate the acceleration of the system:
Net force = ma
ma = Force due to friction + Force due to tension
Applying Newton's second law, we have:
(1.70 kg) × a = (2.0 N) + (3.92 N)
Now, solve for the acceleration (a):
a = [(2.0 N) + (3.92 N)] / (1.70 kg)
a ≈ 4.12 m/s²
Next, we can use the equations of motion to find the distance traveled by the cart in 1.5 seconds. The equation we'll use is:
Distance (d) = Initial velocity (v₀) × time (t) + 0.5 × acceleration (a) × time (t)²
Given that the cart is released from rest (v₀ = 0) and the time is 1.5 seconds, we can plug in the known values:
d = 0 + 0.5 × (4.12 m/s²) × (1.5 s)²
Now, solve for the distance traveled (d):
d ≈ 0.5 × 4.12 m/s² × (1.5 s)²
d ≈ 0.5 × 4.12 m/s² × 2.25 s²
d ≈ 2.06 m
Therefore, the cart will travel approximately 2.06 meters in 1.5 seconds.