A 89.0 N grocery cart is pushed 13.0 m along an aisle by a shopper who exerts a constant horizontal force of 40.0 N. If all frictional forces are neglected and the cart starts from rest, what is the grocery cart's final speed?
net force=mass*acceleration
40-friction= mass* a
find a.
then,
vf^2=2a*distance
8.94
To find the grocery cart's final speed, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass (m) and acceleration (a). In this case, the net force is the force applied by the shopper (F) and the mass of the cart is unknown.
We can start by calculating the acceleration of the cart. We know that the net force is equal to the mass of the cart times the acceleration, so we can rearrange the formula as follows:
a = F/m
In this case, the force applied by the shopper is 40.0 N. Now, since we're neglecting frictional forces, the net force is equal to the applied force. Therefore, we have:
a = 40.0 N/m
Next, we need to find the mass of the cart. To do this, we can use the formula for weight, which is equal to mass times the acceleration due to gravity (g). In this case, the weight of the cart is given as 89.0 N. So we can write:
Weight = mass * g
89.0 N = mass * 9.8 m/s^2
Now, we can calculate the mass of the cart:
mass = 89.0 N / 9.8 m/s^2
Finally, we can use the equation v^2 = u^2 + 2as to find the final velocity (v), where u is the initial velocity (which is 0 m/s since the cart starts from rest), a is the acceleration, and s is the distance traveled.
Plugging in the values, we get:
v^2 = 0^2 + 2 * 40.0 N / m * 13.0 m
Simplifying,
v^2 = 2 * 40.0 N / m * 13.0 m
Now let's substitute the value of mass we calculated earlier:
v^2 = 2 * 40.0 N / (89.0 N / 9.8 m/s^2) * 13.0 m
v^2 = 2 * 40.0 N * 9.8 m/s^2 / 89.0 N * 13.0 m
v^2 = 2 * 40.0 N * 9.8 m/s^2 * 1 / 89.0 * 13.0
v^2 ≈ 0.4747
Taking the square root of both sides to find the final velocity:
v ≈ √0.4747
v ≈ 0.689 m/s
Therefore, the grocery cart's final speed is approximately 0.689 m/s.