an elevator has a weight of 14,700 n and has an acceleration of -0.30 m/s². the free - body diagram shows…

an elevator has a weight of 14,700 n and has an acceleration of -0.30 m/s². the free - body diagram shows the forces acting on the elevator. to the nearest whole number, what is the force of tension, fₜ, acting on the elevator? 450 n 1500 n 14,250 n 15,150 n

an elevator has a weight of 14,700 n and has an acceleration of -0.30 m/s². the free - body diagram shows the forces acting on the elevator. to the nearest whole number, what is the force of tension, fₜ, acting on the elevator? 450 n 1500 n 14,250 n 15,150 n

Answer

Explanation:

Step1: Find the mass of the elevator

We know that weight $F_g = mg$, so $m=\frac{F_g}{g}$. Given $F_g = 14700\ N$ and $g = 9.8\ m/s^2$, then $m=\frac{14700}{9.8}=1500\ kg$.

Step2: Apply Newton's second - law

According to Newton's second - law $F_{net}=ma$. The net force $F_{net}=F_t - F_g$, so $F_t=F_g+ma$. Substitute $F_g = 14700\ N$, $m = 1500\ kg$, and $a=- 0.30\ m/s^2$ into the formula. Then $F_t=14700+1500\times(-0.30)$.

Step3: Calculate the tension force

$F_t=14700 - 450=14250\ N$.

Answer:

C. 14,250 N