Question

a pokémon trainer pushes a bag of pokéballs with a force of 115 n across a horizontal surface. the coefficient of friction between the bag and the surfaces 0.70. the bag accelerates at 4.7 m/s². what is the mass of the bag?
golem uses strength to move a 100 kg boulder. the force he is able to apply is 900 n. psyduck tries to help but instead pushes downwards with a force of 150 n. if the boulder accelerates at a rate of 6.8 m/s², determine the coefficient of friction between the boulder and the ground.
an elevator moving downwards at 1.4 m/s slows to a stop over a distance of 3.0 m. a 65 kg person in the elevator stands on a scale which shows force in newtons.
a) what does the scale read?
b) compared to when the elevator is at rest, is the scale reading greater, lower or the same? use principles of physics to explain your answer.

Explanation:

Step1: Find the net - force formula for the first problem

According to Newton's second law $F_{net}=ma$. The net - force acting on the bag of Pokéballs is $F - F_f=ma$, where $F$ is the applied force, $F_f=\mu N$ is the frictional force, and on a horizontal surface $N = mg$. So $F-\mu mg=ma$.

Step2: Solve for the mass $m$ in the first problem

Rearrange the equation $F-\mu mg=ma$ to get $F = m(a+\mu g)$. Then $m=\frac{F}{a + \mu g}$. Substitute $F = 115\ N$, $\mu=0.70$, $a = 4.7\ m/s^2$, and $g = 9.8\ m/s^2$ into the formula:
\[m=\frac{115}{4.7+0.7\times9.8}=\frac{115}{4.7 + 6.86}=\frac{115}{11.56}\approx10\ kg\]

Step3: Find the net - force formula for the second problem

For the boulder, the net - force is $F_{net}=F - F_f=ma$. The normal force $N=mg + F_{down}$, where $F$ is the applied force to move the boulder, $F_f=\mu N$, $m = 100\ kg$, $F = 900\ N$, $F_{down}=150\ N$, and $a = 6.8\ m/s^2$. First, find the net - force $F_{net}=ma=100\times6.8 = 680\ N$. Then the frictional force $F_f=F - F_{net}=900 - 680=220\ N$. The normal force $N=mg+F_{down}=100\times9.8+150=980 + 150 = 1130\ N$.

Step4: Solve for the coefficient of friction $\mu$ in the second problem

Since $F_f=\mu N$, then $\mu=\frac{F_f}{N}$. Substitute $F_f = 220\ N$ and $N = 1130\ N$ into the formula: $\mu=\frac{220}{1130}\approx0.195$.

Step5: Find the acceleration of the elevator in the third problem

Use the kinematic equation $v^2=v_0^2+2ad$. Here, $v = 0\ m/s$, $v_0=- 1.4\ m/s$ (downward is negative), and $d=-3.0\ m$ (downward is negative). So $0=( - 1.4)^2+2a(-3.0)$. Then $0 = 1.96-6a$, and $6a = 1.96$, $a=\frac{1.96}{6}\approx0.33\ m/s^2$ (the positive value means the acceleration is upward).

Step6: Find the normal force (scale reading) in the third problem

For the person in the elevator, according to Newton's second law $F_{net}=N - mg=ma$ (upward is positive). Then $N=m(g + a)$. Substitute $m = 65\ kg$, $g = 9.8\ m/s^2$, and $a = 0.33\ m/s^2$ into the formula: $N=65\times(9.8 + 0.33)=65\times10.13 = 658.45\ N$.

Step7: Answer part b of the third problem

When the elevator is at rest, $N_0=mg=65\times9.8 = 637\ N$. Since $N = 658.45\ N$ and $N_0 = 637\ N$, the scale reading is greater. This is because when the elevator is slowing down while moving downwards, there is an upward acceleration. According to Newton's second law, the normal force (scale reading) must be greater than the weight of the person to provide the net - upward force for deceleration.

Answer:

1. The mass of the bag is approximately $10\ kg$.
2. The coefficient of friction between the boulder and the ground is approximately $0.195$.
3. a) The scale reads $658.45\ N$.

b) The scale reading is greater. When the elevator is slowing down while moving downwards, there is an upward acceleration. According to Newton's second law, the normal force (scale reading) must be greater than the weight of the person to provide the net - upward force for deceleration.