If a cart and rider are pulled with a constant force of 74.0 N, and the friction force is 29.0 N, how fast will the cart and rider be traveling after 8.70 seconds?
Thanks a lot!
the acceleration is 0.31 m/s2
(1/2) a t^2, where a = acceleration
The data on forces is not needed.
You asked for the velocity. That will ewqual at
In my previous response I gave you the distance travelled.
To determine how fast the cart and rider will be traveling after 8.70 seconds, we need to apply Newton's second law of motion. This law states that the net force acting on an object is equal to the product of its mass and acceleration.
First, let's calculate the net force acting on the cart and rider. The net force can be found by subtracting the friction force from the applied force:
Net force = Applied force - Friction force
Net force = 74.0 N - 29.0 N
Net force = 45.0 N
Next, we can use the net force and the mass of the cart and rider to calculate the acceleration. The acceleration can be found using the equation:
Net force = mass × acceleration
Rearranging the equation, we get:
Acceleration = Net force / mass
Assuming we have the mass of the cart and rider, we can substitute the values and calculate the acceleration.
Once we have the acceleration, we can use it to find the final velocity of the cart and rider after 8.70 seconds. The equation relating final velocity, initial velocity, acceleration, and time is given by:
Final velocity = Initial velocity + (acceleration × time)
Since the initial velocity is not given in the problem, we'll assume that the cart starts from rest, meaning the initial velocity is zero.
Now, let's calculate the acceleration and the final velocity using the given values.