Question

1. a 40 kg cart is pushed with a 200 n force. if friction is 40 n, what is the acceleration?
2. a 20 kg box is pulled with 60 n, but friction opposes motion with 20 n. find the acceleration.
3. a 2,000 kg truck accelerates at 1.5 m/s². what net force acts on the truck?
4. a 0.25 kg baseball is hit with a force of 20 n. what acceleration does it experience?
5. a 70 kg skateboarder accelerates down a hill at 1.2 m/s². find the net force acting on the skateboarder.
6. a 3 kg cart is pushed by a 20 n force while friction is 2 n. determine its acceleration.

Explanation:

Step1: Calculate net force

According to Newton's second law \(F_{net}=F - f\). For problem 11, \(F = 200N\), \(f=40N\), so \(F_{net}=200 - 40=160N\).

Step2: Calculate acceleration

Using \(a=\frac{F_{net}}{m}\). For problem 11, \(m = 40kg\), \(a=\frac{160}{40}=4m/s^{2}\).

For problem 12:

Step1: Calculate net force

\(F_{net}=60 - 20 = 40N\)

Step2: Calculate acceleration

\(m = 20kg\), \(a=\frac{40}{20}=2m/s^{2}\)

For problem 13:

Step1: Calculate net force

Using \(F_{net}=ma\). \(m = 2000kg\), \(a = 1.5m/s^{2}\), \(F_{net}=2000\times1.5 = 3000N\)

For problem 14:

Step1: Calculate acceleration

\(m=0.25kg\), \(F = 20N\), \(a=\frac{20}{0.25}=80m/s^{2}\)

For problem 15:

Step1: Calculate net force

\(m = 70kg\), \(a = 1.2m/s^{2}\), \(F_{net}=70\times1.2=84N\)

For problem 16:

Step1: Calculate net force

\(F_{net}=20 - 2=18N\)

Step2: Calculate acceleration

\(m = 3kg\), \(a=\frac{18}{3}=6m/s^{2}\)

Answer:

1. \(4m/s^{2}\)
2. \(2m/s^{2}\)
3. \(3000N\)
4. \(80m/s^{2}\)
5. \(84N\)
6. \(6m/s^{2}\)