Question

chapter 5
question 6 (1 point)
a brick is moving at a speed of 3 m/s and a pebble is moving at a speed of 5 m/s. if both objects have the same kinetic energy, what is the ratio of the bricks mass to the rocks mass?
25 to 9
5 to 3
12.5 to 4.5
3 to 5
question 7 (1 point)
a 500-kg elevator is pulled upward with a constant force of 5500 n for a distance of 50.0 m. what is the work done by the weight of the elevator?
2.75 × 10^5 j
-2.45 × 10^5 j
3.00 × 10^4 j
-5.20 × 10^5 j

Explanation:

Question 6

Step1: Recall Kinetic Energy Formula

The kinetic energy formula is $KE = \frac{1}{2}mv^2$, where $m$ is mass and $v$ is speed. Let the brick's mass be $m_b$, speed $v_b = 3\ m/s$; pebble's (rock's) mass be $m_p$, speed $v_p = 5\ m/s$. Given $KE_b = KE_p$, so $\frac{1}{2}m_bv_b^2=\frac{1}{2}m_pv_p^2$.

Step2: Simplify the Equation

Cancel $\frac{1}{2}$ from both sides: $m_bv_b^2 = m_pv_p^2$. Rearrange to find $\frac{m_b}{m_p}=\frac{v_p^2}{v_b^2}$.

Step3: Substitute Values

Substitute $v_b = 3$, $v_p = 5$: $\frac{m_b}{m_p}=\frac{5^2}{3^2}=\frac{25}{9}$.

Step1: Recall Work Formula

Work done by a force is $W = Fd\cos\theta$, where $F$ is force, $d$ is distance, $\theta$ is angle between force and displacement. Weight $F_g = mg$, $m = 500\ kg$, $g = 9.8\ m/s^2$, so $F_g = 500\times9.8 = 4900\ N$. Displacement $d = 50.0\ m$, angle between weight (downward) and displacement (upward) is $\theta = 180^\circ$, $\cos180^\circ=-1$.

Step2: Calculate Work Done by Weight

$W = F_gd\cos\theta = 4900\times50.0\times(-1)= -245000\ J = -2.45\times10^5\ J$.

Answer:

25 to 9

Question 7