A 2 kg lab cart is sliding across across a horizontal frictionless surface at a constant velocity of 4.0 meters per second east. What will be the cart's velocity after a 6.0-newton westward force acts on it for 2.0 seconds and in what direction?
Update 2: what i don't get is i did force=(m*change in velocity)/change in time, so 6=[2(v final-4)/2 and got 10.
Update 3: but its not right
Update 4: my teacher used m(v final- v initial)=F*change in time. why is that?
-->J = Impulse = F * t = change in momentum
--> Units for Impulse can be N*s or kg*m/s
--> p = momentum = m * v units are kg*m/s
--> East is +, West is -
The change in momentum is = m(v final- v initial)
Impulse = F * t so...
m(v final- v initial) = F * t
(2kg * v final) - (2kg * 4 m/s) = -[6N * 2s]
(2kg*v) - 8 kg m/s = -12 kg m/s
+8 kg m/s +8 kg m/s
(2kg*v) =-4 kg m/s
v= -2 m/s (the negative signifies West)
Answer=2 m/s West
To solve this problem, we can use Newton's second law of motion, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration. In this case, since the velocity of the cart is constant, the acceleration is zero.
Let's break down the problem step by step:
1. Initially, the cart is sliding with a constant velocity of 4.0 meters per second east. Since there is no opposing force or friction, we can assume the net force acting on the cart is zero.
2. When a 6.0-newton westward force acts on the cart for 2.0 seconds, it will cause the cart to experience an acceleration in the opposite direction (westward).
3. In order to find the new velocity of the cart, we need to calculate its acceleration first. Using Newton's second law, we have:
Force = mass * acceleration
6.0 N = 2 kg * acceleration
acceleration = 3.0 m/s^2 (westward)
4. Now, to find the final velocity of the cart, we can use the following kinematic equation:
v_final = v_initial + (acceleration * time)
v_final = 4.0 m/s + (3.0 m/s^2 * 2.0 s)
v_final = 4.0 m/s + 6.0 m/s
v_final = 10.0 m/s (eastward)
So, the final velocity of the cart after the 6.0-newton westward force acts on it for 2.0 seconds is 10.0 meters per second eastward.
Your teacher's equation, m(v final - v initial) = F * change in time, is the correct equation to use in this scenario. Let's break it down to understand why:
Given information:
- Mass of cart (m): 2 kg
- Initial velocity of cart (v initial): 4.0 m/s east
- Force acting on the cart (F): 6.0 N to the west
- Time duration (change in time): 2.0 seconds
Using Newton's second law, the equation is written as F = m * a, where F is the force, m is the mass, and a is the acceleration. In this case, we assume that the force is constant, so the acceleration will also be constant.
First, let's calculate the acceleration:
F = m * a
6.0 N = 2 kg * a
a = 6.0 N / 2 kg
a = 3.0 m/s^2 to the west
Now, using the kinematic equation, v final = v initial + a * change in time, we can find the final velocity:
v final = v initial + a * change in time
v final = 4.0 m/s + 3.0 m/s^2 * 2.0 s
v final = 4.0 m/s + 6.0 m/s
v final = 10.0 m/s east
Therefore, after the 6.0-newton westward force acts on the cart for 2.0 seconds, the cart's final velocity will be 10.0 m/s east.