Question

1. mr. cs students pull a cart that has a mass of 12 kg. they applied a force of 4n to the cart. calculate the carts acceleration.
2. ms. martinez pushed mr. e (who has a mass of 200kg). he fell with an acceleration of 2 m/s². how much force did ms. martinez apply to mr. e?
3. draw a picture to help find the answer. sue pulls on the right side of a rope with a force of 4n. jim pulls on the left side of the rope with a force of 10n. which way will the rope move and with how much net force?

Explanation:

Problem 8

Step1: Recall Newton's second law

Newton's second law is \(F = ma\), where \(F\) is force, \(m\) is mass, and \(a\) is acceleration. We need to solve for \(a\), so we can rewrite the formula as \(a=\frac{F}{m}\).

Step2: Substitute the given values

Given \(F = 4N\) and \(m=12kg\). Substitute into \(a=\frac{F}{m}\), we get \(a=\frac{4}{12}\).

Step3: Simplify the fraction

\(\frac{4}{12}=\frac{1}{3}\approx0.33\) (rounded to two decimal places)

Step1: Use Newton's second law

Newton's second law \(F = ma\). Here, \(m = 200kg\) and \(a = 2m/s^{2}\)

Step2: Substitute values into the formula

Substitute \(m\) and \(a\) into \(F=ma\), we have \(F=(200)(2)\)

Step3: Calculate the result

\((200)(2)=400\)

Step1: Determine the direction of net - force

Forces are in opposite directions. Let's assume the left - direction is positive. The force from Jim (\(F_J = 10N\)) is in the positive (left) direction and the force from Sue (\(F_S=4N\)) is in the negative (right) direction.

Step2: Calculate the net force

The net force \(F_{net}=F_J - F_S\). Substitute \(F_J = 10N\) and \(F_S = 4N\), we get \(F_{net}=10 - 4\)

Step3: Analyze the result

\(F_{net}=6N\). Since \(F_{net}>0\), the rope moves to the left.

Answer:

The cart's acceleration is \(\frac{1}{3}m/s^{2}\) or approximately \(0.33m/s^{2}\)

Problem 9