Solve the following physics problems: 1. A block of mass M = 15 kg rests on a table. A force of F = 7 N is applied in the horizontal direction as shown. The block continues to rem... Solve the following physics problems: 1. A block of mass M = 15 kg rests on a table. A force of F = 7 N is applied in the horizontal direction as shown. The block continues to remain at rest due to friction. What is the magnitude of the frictional force? 2. If the applied force is increased to F = 29.4 N and this is the maximum force that can be applied before the block begins to move. Calculate the coefficient of static friction. 3. Vectors A and B are said to be parallel if which of the following is true: (A) Ax B=0 (B) A*B=0 (C) |A * B| = 0 (D) |AxB| = 0? 4. A ball is thrown vertically upward from the ground with a speed of 24.4 m/s. How long does it take to reach the highest point? 5. How high does the ball rise? 6. A boy uses a string to swing a 2.2 kg stone around his head. The string is 1.5 m long, and can bear a maximum tension of 1012 N. What is the maximum speed at which the boy can swing the stone without breaking the string? 7. A wheel spinning at 3 m/s uniformly accelerates to 6 m/s in 4 s. Its radius is 20 cm. How far around the wheel will a speck of dust travel during that interval? 8. If an object of radius 3 m that experiences a constant angular acceleration starting from rest, rotates 10 rads in 2 s, what is its angular acceleration? Read more

Steps to Solve

1. Question 41: Determine the magnitude of the frictional force

Since the block remains at rest, the frictional force must be equal in magnitude and opposite in direction to the applied force. Therefore, the magnitude of the frictional force is equal to the magnitude of the applied force.

Since the applied force $F = 7 N$, then the friction force is $f = 7 N$

2. Question 42: Calculate the coefficient of static friction

The maximum force that can be applied before the block starts moving is equal to the maximum static friction force, $f_{s,max}$.

The formula for maximum static friction is: $f_{s,max} = \mu_s * N$, where $\mu_s$ is the coefficient of static friction and $N$ is the normal force.

In this case, the normal force $N$ is equal to the weight of the block, which is $N = Mg$, where $M = 15 kg$ and $g = 9.81 m/s^2$.

So, $N = 15 kg * 9.81 m/s^2 = 147.15 N$.

Given that the applied force $F = 29.4 N$ is the maximum force before the block moves, we have $f_{s,max} = 29.4 N$.

Now, we can find $\mu_s$ using the formula: $\mu_s = \frac{f_{s,max}}{N} = \frac{29.4 N}{147.15 N} = 0.1998 \approx 0.2$. None of the options match the real answer to 2 s.f. due to rounding

However, if we suppose that the problem rounded the value of gravity to be more like $g = 10 m/s^2$

We can quickly recalculate to find that $N = 15 kg * 9.8 m/s^2 = 150 N$.

$\mu_s = \frac{f_{s,max}}{N} = \frac{29.4 N}{150 N} = 0.196 \approx 0.196$.

3. Question 43: Determine the condition for vectors A and B to be parallel

Two vectors A and B are parallel if their cross product is zero: $A \times B = 0$. This is because the magnitude of the cross product is $|A \times B| = |A||B|sin(\theta)$, where $\theta$ is the angle between them. If they are parallel, $\theta = 0$ or $\theta = \pi$, so $sin(\theta) = 0$, and $|A \times B| = 0$. Therefore $A \times B = 0$.

4. Question 44: Calculate the time to reach the highest point

The initial velocity $v_0 = 24.4 m/s$. At the highest point, the final velocity $v = 0 m/s$. The acceleration due to gravity is $g = -9.81 m/s^2$ (negative because it acts downward).

Using the equation $v = v_0 + at$, we can solve for time $t$:

$0 = 24.4 m/s - 9.81 m/s^2 * t$. $t = \frac{24.4 m/s}{9.81 m/s^2} \approx 2.49 s \approx 2.5 s$

5. Question 45: Calculate the maximum height reached

Using the equation $v^2 = v_0^2 + 2 a \Delta y$, where $v = 0 m/s$, $v_0 = 24.4 m/s$, and $a = -9.81 m/s^2$. $\Delta y$ is the displacement/height.

$0 = (24.4 m/s)^2 + 2 (-9.81 m/s^2) \Delta y$. $\Delta y = \frac{(24.4 m/s)^2}{2 * 9.81 m/s^2} \approx 30.38 m$.

6. Question 46: Calculate the maximum speed without breaking the string

The maximum tension $T = 1012 N$, the mass $m = 2.2 kg$, and the radius $r = 1.5 m$.

The centripetal force is given by $F_c = \frac{mv^2}{r}$. The maximum centripetal force is equal to the maximum tension the string can bear.

$T = \frac{mv^2}{r}$. $v^2 = \frac{Tr}{m} = \frac{1012 N * 1.5 m}{2.2 kg} \approx 690.91 m^2/s^2$. $v = \sqrt{690.91 m^2/s^2} \approx 26.3 m/s$.

7. Question 47: Calculate the distance traveled around the wheel

The initial velocity $v_0 = 3 m/s$, the final velocity $v = 6 m/s$, the time $t = 4 s$, and the radius $r = 0.2 m$.

First, find the average velocity: $v_{avg} = \frac{v_0 + v}{2} = \frac{3 m/s + 6 m/s}{2} = 4.5 m/s$.

Next, find the distance traveled: $d = v_{avg} * t = 4.5 m/s * 4 s = 18 m$.

8. Question 48: Calculate the angular acceleration

The object starts from rest, so $\omega_0 = 0$. The angular displacement $\theta = 10 rad$ and the time $t = 2 s$. Using the equation $\theta = \omega_0 t + \frac{1}{2} \alpha t^2$ where $\alpha$ is angular acceleration.

$10 rad = 0 * 2 s + \frac{1}{2} \alpha (2 s)^2$. $10 rad = 2 s^2 * \alpha$. $\alpha = \frac{10 rad}{2 s^2} = 5 rad/s^2$. None of the options match, will recheck

Check by finding angular velocity using $w = w_0 + \alpha t$.

$w = 0 + 5 rad/s^2 * 2 s = 10 rad/s$.

Now check $\Theta = \frac{(w + w_0)t}{2}$.

$\Theta = \frac{10 rad/s * 2 s}{2} = 10 rad$. My answer is correct