Question

part 2 word problems: make sure to show all work and circle/box your answer so that it stands out. double check all units! (3 pts each)

1. a 98.0 kg man is accelerating forward at a rate of 1.13 m/s². what is the unbalanced force necessary to produce this movement?
2. a 5.5 kg box is pushed across the floor. the net force applied to the box is 9.7 n. what is the acceleration of the box?
3. what is the mass of an object that has an acceleration of 2.63 m/s² when an unbalanced force of 112 n is applied to it?
4. if a bicyclist has a mass of 72.0 kg and a velocity of 15.0 m/s, what is the momentum of the bicyclist?
5. if a truck has 40,500 kg*m/s of momentum and is moving with a velocity of 91.0 m/s, what is the truck’s mass?

bonus: a runner has a speed of 6 m/s. they see the finish line and speed up to 8 m/s. this happens in 5 seconds. if the runner has a mass of 75 kg, with what force did the runner cross the finish line? show all work to receive full credit.

Explanation:

11. Force Calculation

Step1: Identify the formula

Use Newton's second law \(F = ma\), where \(m = 98.0\space kg\) and \(a=1.13\space m/s^{2}\)

Step2: Substitute values

\(F=(98.0\space kg)\times(1.13\space m/s^{2})\)

Step3: Calculate

\(F = 110.74\space N\)

12. Acceleration Calculation

Step1: Rearrange the formula

From \(F = ma\), we get \(a=\frac{F}{m}\), with \(F = 9.7\space N\) and \(m = 5.5\space kg\)

Step2: Substitute values

\(a=\frac{9.7\space N}{5.5\space kg}\)

Step3: Calculate

\(a\approx1.76\space m/s^{2}\)

13. Mass Calculation

Step1: Rearrange the formula

From \(F = ma\), we get \(m=\frac{F}{a}\), with \(F = 112\space N\) and \(a = 2.63\space m/s^{2}\)

Step2: Substitute values

\(m=\frac{112\space N}{2.63\space m/s^{2}}\)

Step3: Calculate

\(m\approx42.6\space kg\)

14. Momentum Calculation

Step1: Use the formula

Momentum \(p=mv\), with \(m = 72.0\space kg\) and \(v = 15.0\space m/s\)

Step2: Substitute values

\(p=(72.0\space kg)\times(15.0\space m/s)\)

Step3: Calculate

\(p = 1080\space kg\cdot m/s\)

15. Mass from Momentum

Step1: Rearrange the formula

From \(p = mv\), we get \(m=\frac{p}{v}\), with \(p = 40500\space kg\cdot m/s\) and \(v = 91.0\space m/s\)

Step2: Substitute values

\(m=\frac{40500\space kg\cdot m/s}{91.0\space m/s}\)

Step3: Calculate

\(m\approx445\space kg\)

Bonus: Force Calculation (Using Acceleration)

Step1: Calculate acceleration

Use \(a=\frac{v - u}{t}\), where \(u = 6\space m/s\), \(v = 8\space m/s\), \(t = 5\space s\)
\(a=\frac{8 - 6}{5}=\frac{2}{5}=0.4\space m/s^{2}\)

Step2: Use Newton's second law

\(F = ma\), with \(m = 75\space kg\) and \(a = 0.4\space m/s^{2}\)
\(F=(75\space kg)\times(0.4\space m/s^{2})\)

Step3: Calculate

\(F = 30\space N\)

Answer:

1. \(110.74\space N\)
2. \(1.76\space m/s^{2}\)
3. \(42.6\space kg\)
4. \(1080\space kg\cdot m/s\)
5. \(445\space kg\)

Bonus: \(30\space N\)