Question

1. (a) the mass of this car is 970 kg. calculate the acceleration of the car. opposing force = 560 n 275 n 395 n options: 106,700 m/s^2; 8.8 m/s^2; 0.1 m/s^2

Explanation:

Step1: Calculate net force

First, find the total applied force by adding 275 N and 395 N. Then subtract the opposing force (560 N) to get the net force.
Total applied force: \(275 + 395 = 670\) N
Net force \(F_{net} = 670 - 560 = 110\) N

Step2: Use Newton's second law (\(F = ma\)) to find acceleration

Newton's second law is \(F = ma\), so rearrange to \(a=\frac{F_{net}}{m}\). Given \(m = 970\) kg and \(F_{net}=110\) N.
\(a=\frac{110}{970}\approx0.113\) m/s² (approx 0.1 m/s² when rounded, but let's check the calculation again for accuracy. Wait, maybe I miscalculated the net force. Wait, 275 + 395 is 670, minus 560 is 110. Then 110/970 ≈ 0.113, but the options have 0.1 m/s². Wait, maybe the forces: let's recheck. The applied forces are 275 N and 395 N (both in the same direction?), and opposing force is 560 N. So net force \(F = (275 + 395) - 560 = 670 - 560 = 110\) N. Then acceleration \(a = F/m = 110 / 970 ≈ 0.113\) m/s², which is approximately 0.1 m/s² (maybe due to rounding in the problem). Wait, but let's check the other option: 8.8 m/s² would be if net force was 970*8.8≈8536 N, which is too big. 106700 is way too big. So the correct net force leads to ~0.1 m/s². Wait, maybe I made a mistake in force directions. Wait, the diagram: 275 N and 395 N are applied forces (maybe in the same direction), opposing force is 560 N. So net force is sum of applied minus opposing. So 275 + 395 = 670; 670 - 560 = 110. Then 110 / 970 ≈ 0.113, so ~0.1 m/s².

Answer:

0.1 m/s² (corresponding to the option with 0.1 m/s²)