Question

1. what is the kinetic energy of a 0.01 kg bullet traveling at a velocity of 700 m/s?

a. 3.5 j
b. 7 j
c. $2.45 \times 10^3$ j
d. $2.45 \times 10^5$ j

1. a marble rolling across a flat, hard surface at 2 m/s rolls up a ramp. assuming that $g = 10$ m/s² and no energy is lost to friction, what will be the vertical height of the marble when it comes to a stop before rolling back down? ignore effects due to the rotational kinetic energy.

a. 0.1 m
b. 0.2 m
c. 0.4 m
d. 2 m

Explanation:

Question 16

Step1: Recall Kinetic Energy Formula

The formula for kinetic energy (KE) is $KE = \frac{1}{2}mv^2$, where $m$ is mass and $v$ is velocity.

Step2: Substitute Values

Given $m = 0.01\space kg$ and $v = 700\space m/s$. Substitute into the formula:
$KE=\frac{1}{2}\times0.01\space kg\times(700\space m/s)^2$
First, calculate $(700)^2 = 490000$. Then, $\frac{1}{2}\times0.01 = 0.005$. Multiply: $0.005\times490000 = 2450\space J = 2.45\times10^3\space J$.

Step1: Use Energy Conservation

Kinetic energy (KE) converts to gravitational potential energy (PE) at the top. So, $KE = PE$.
KE formula: $\frac{1}{2}mv^2$, PE formula: $mgh$.
Set $\frac{1}{2}mv^2 = mgh$. The mass $m$ cancels out.

Step2: Solve for Height $h$

Simplify: $\frac{1}{2}v^2 = gh$. Given $v = 2\space m/s$, $g = 10\space m/s^2$.
$h=\frac{v^2}{2g}=\frac{(2)^2}{2\times10}=\frac{4}{20}=0.2\space m$.

Answer:

c. $2.45\times 10^3$ J

Question 17