Question

1. given the position versus time graph below, draw corresponding velocity vs time and acceleration vs time graphs.
2. a worker at amazon places a small box with a mass of “m” kilograms on a platform scale. the worker ties a string to the box and gently pulls up on the box. the scale measures a non - zero force and the box remains motionless.

a. draw a fbd
b. write an equilibrium equation
c. what would be the reading on the scale? solve the equilibrium equation for the reading on the scale. hint: there are no numbers to plug in here, just move the variables around.

1. a cat hangs at rest suspended from two ropes in a tree. the cat has a mass of 4 kg. the rope pulling to the right pulls on the cat with a tension force of 78.4 newtons. determine the tension force in the rope that is at an angle, and determine the angle θ.
2. a ball starts at rest on a ramp and moves 3 meters in 4 seconds.

a. calculate the acceleration of the ball
b. calculate the final velocity of the ball at the base of the ramp

1. a ball is kicked upwards at a velocity of v₀. determine the maximum height of the ball in terms of v₀ and g.
2. a nerf dart is fired horizontally from an initial height of 2.2 meters off of the ground. the dart travels 7.8 meters before it hits the ground. calculate the initial velocity of the nerf dart in the x direction.

Explanation:

Step 1: Solve problem 9

The slope of the position - time graph gives velocity. At the start, the slope is negative and large, then it decreases to zero at the minimum of the position - time graph, and then it becomes positive and increases. So, for the velocity - time graph, it will start at a negative value, decrease to zero, and then increase to a positive value. The slope of the velocity - time graph gives acceleration. The acceleration is positive when the velocity is increasing (concave - up on the position - time graph) and negative when the velocity is decreasing (concave - down on the position - time graph). So, the acceleration - time graph will be positive, then zero at the inflection point (where velocity is at a minimum or maximum), and then positive again. However, since we are not asked to provide numerical values but just the general shape of the graphs, we can't provide a more detailed numerical solution for this part.

Step 2: Solve problem 10a

A free - body diagram (FBD) of the box on the scale has three forces: the force of gravity $F_g=mg$ acting downwards, the normal force $N$ (the reading on the scale) acting upwards, and the tension force $T$ applied by the worker pulling upwards.

Step 3: Solve problem 10b

The equilibrium equation for the vertical forces is $\sum F_y = 0$. So, $N+T - mg=0$.

Step 4: Solve problem 10c

We solve the equilibrium equation $N+T - mg = 0$ for $N$. Rearranging the equation gives $N=mg - T$.

Step 5: Solve problem 11

The weight of the cat $W = mg$, where $m = 4$ kg and $g=9.8$ m/s², so $W=4\times9.8 = 39.2$ N. Let the tension in the right - hand rope be $T_1 = 78.4$ N and the tension in the angled rope be $T_2$. In the vertical direction, $T_2\sin\theta=mg$ and in the horizontal direction $T_2\cos\theta=T_1$. Dividing the vertical equation by the horizontal equation gives $\tan\theta=\frac{mg}{T_1}$. Substituting $mg = 39.2$ N and $T_1 = 78.4$ N, we get $\tan\theta=\frac{39.2}{78.4}=0.5$, so $\theta=\arctan(0.5)\approx26.6^{\circ}$. Then, from $T_2\sin\theta=mg$, we can find $T_2=\frac{mg}{\sin\theta}=\frac{39.2}{\sin(26.6^{\circ})}\approx88.9$ N.

Step 6: Solve problem 12a

We use the kinematic equation $x = v_0t+\frac{1}{2}at^{2}$. Since $v_0 = 0$ (starts at rest), $x = 3$ m and $t = 4$ s. Substituting into the equation $x=\frac{1}{2}at^{2}$, we solve for $a$:
\[a=\frac{2x}{t^{2}}=\frac{2\times3}{4^{2}}=\frac{6}{16}=0.375\ m/s^{2}\]

Step 7: Solve problem 12b

We use the kinematic equation $v = v_0+at$. Since $v_0 = 0$ and $a = 0.375$ m/s² and $t = 4$ s, then $v=0 + 0.375\times4=1.5$ m/s.

Step 8: Solve problem 13

At the maximum height, the final velocity $v = 0$. We use the kinematic equation $v^{2}-v_0^{2}=- 2gh$. Rearranging for $h$ gives $h=\frac{v_0^{2}}{2g}$.

Step 9: Solve problem 14

In the vertical direction, the dart is in free - fall. The initial vertical velocity $v_{0y}=0$ m/s, the acceleration $a = g=9.8$ m/s² and the vertical displacement $y=- 2.2$ m. Using the equation $y = v_{0y}t+\frac{1}{2}at^{2}$, we have $y=\frac{1}{2}at^{2}$. Solving for the time of flight $t$:
\[t=\sqrt{\frac{-2y}{a}}=\sqrt{\frac{-2\times(-2.2)}{9.8}}\approx0.67\ s\]
In the horizontal direction, $x = v_{0x}t$. Given $x = 7.8$ m and $t\approx0.67$ s, we solve for $v_{0x}$:
\[v_{0x}=\frac{x}{t}=\frac{7.8}{0.67}\approx11.64\ m/s\]

Answer:

For problem 9: Sketch general - shape velocity and acceleration vs time graphs as described above.
For problem 10a: FBD with $F_g$ down, $N$ up, $T$ up.
For problem 10b: $N+T - mg = 0$
For problem 10c: $N=mg - T$
For problem 11: $T_2\approx88.9$ N, $\theta\approx26.6^{\circ}$
For problem 12a: $a = 0.375\ m/s^{2}$
For problem 12b: $v = 1.5$ m/s
For problem 13: $h=\frac{v_0^{2}}{2g}$
For problem 14: $v_{0x}\approx11.64\ m/s$