A 1.5-kg cart is pulled with a force of 7.3 N at an angle of 40° above the horizontal. If a kinetic friction force of 3.2 N acts against the motion, the cart’s acceleration along the horizontal surface will be
net force=ma
7.3cos40-3.2=ma
solve for a
Well, well, well, looks like we have ourselves a physics problem here. Let's see what we can do!
Given that the cart is being pulled with a force of 7.3 N at an angle of 40° above the horizontal, we can break down the force into its horizontal and vertical components. The horizontal component of the force can be found using some good old trigonometry.
So, the horizontal force pulling the cart would be 7.3 N * cos(40°), which gives us 5.56 N (approximately).
Now, let's talk about that pesky kinetic friction force of 3.2 N acting against the motion. You know, friction is like that annoying friend who always tries to ruin the fun. In this case, it's trying to slow down the cart's motion.
So, the net force acting on the cart along the horizontal surface would be the force pulling the cart (5.56 N) minus the kinetic friction force (3.2 N), which gives us a net force of 2.36 N (approximately).
And finally, we can use good ol' Newton's second law (F = ma) to find the acceleration. Plugging in the values, we get:
2.36 N = 1.5 kg * a
Solving for a, we find that the cart's acceleration along the horizontal surface will be approximately 1.57 m/s².
Phew, that was a wild ride, wasn't it? I sure hope this answer didn't pull you off balance with all those numbers and calculations. But hey, who said physics couldn't be fun?
To find the acceleration of the cart along the horizontal surface, we will first resolve the force applied and the force of friction into their horizontal components. Then, we can use Newton's second law of motion to calculate the acceleration.
Step 1: Resolve the applied force into horizontal and vertical components:
The horizontal component of the applied force can be found by multiplying the magnitude of the force by the cosine of the angle:
F_horizontal = 7.3 N * cos(40°)
Step 2: Calculate the net force acting on the cart:
The net force is the difference between the horizontal component of the applied force and the force of friction:
Net force = F_horizontal - Force of friction
Step 3: Determine the acceleration using Newton's second law of motion:
Newton's second law states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration:
Net force = mass * acceleration
Rearranging the equation to solve for acceleration, we get:
acceleration = Net force / mass
Step 4: Substitute the given values to find the acceleration:
acceleration = (F_horizontal - Force of friction) / mass
acceleration = (F_horizontal - 3.2 N) / 1.5 kg
Now, let's calculate the values.
First, calculate the horizontal component of the applied force:
F_horizontal = 7.3 N * cos(40°)
Next, substitute and calculate the net force:
Net force = F_horizontal - 3.2 N
Finally, calculate the acceleration:
acceleration = Net force / mass
Please provide the value for the mass of the cart.
To find the cart's acceleration along the horizontal surface, we can use Newton's second law of motion which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.
We need to resolve the force into its horizontal and vertical components. The horizontal component of the force acting on the cart is given by F_h = F * cos(theta), where F is the applied force and theta is the angle above the horizontal.
F_h = 7.3 N * cos(40°)
F_h ≈ 7.3 N * 0.766
F_h ≈ 5.59 N
Since there is kinetic friction acting against the motion, we need to subtract the friction force from the horizontal force:
F_net = F_h - friction
F_net = 5.59 N - 3.2 N
F_net = 2.39 N
Now we can find the acceleration by rearranging the equation F_net = m * a, where m is the mass of the cart (1.5 kg) and a is the acceleration.
2.39 N = 1.5 kg * a
Solving for a, we get:
a = 2.39 N / 1.5 kg
a ≈ 1.59 m/s^2
Therefore, the cart's acceleration along the horizontal surface will be approximately 1.59 m/s^2.