james pushed his 10 kg cart with a constant horizontal force of 9.0 Newtons over a distance of 3.1 meters. If all frictional forces are neglected and the cart started from rest, what is the grocery cart's final speed in meters per second?
Work done = kinetic energy increase
9.0 * 3.1 = (1/2)*M*V^2 = (0.5)*10*V^2
Solve for V
To determine the final speed of the grocery cart, we can use Newton's second law of motion, which states that the force applied to an object is equal to the product of its mass and acceleration.
Given:
Mass of the cart (m) = 10 kg
Force applied (F) = 9.0 N
Distance traveled (d) = 3.1 m
We need to find the acceleration (a) of the cart using the formula:
F = m * a
Since the cart is initially at rest and there are no frictional forces, the entire force applied will be used to accelerate the cart. Therefore, we can equate the force applied to the mass-acceleration equation:
9.0 N = 10 kg * a
Now, solve for acceleration (a):
a = 9.0 N / 10 kg
a ≈ 0.9 m/s²
Now that we have the acceleration, we can use another equation of motion to find the final speed of the cart. The equation is:
v² = u² + 2ad
Where:
v = final velocity (speed) of the cart
u = initial velocity (which is 0 in this case)
a = acceleration
d = distance traveled
Plugging in the values:
v² = 0² + 2 * 0.9 m/s² * 3.1 m
v² ≈ 5.58 m²/s²
Finally, take the square root of both sides to find the final speed:
v ≈ √5.58 m²/s²
v ≈ 2.36 m/s
Therefore, the grocery cart's final speed is approximately 2.36 meters per second.