a stuntman drives a car with a mass of 1600 kg on a drawbridge. the car accelerates with a constant force of…

a stuntman drives a car with a mass of 1600 kg on a drawbridge. the car accelerates with a constant force of 8000 n. while he is driving, the drawbridge is raised to an incline of 30°. what is the cars new acceleration on this incline, ignoring the force due to air resistance? (recall that (g = 9.8 m/s^{2}))\nforce accelerating car\n30°\nweight\na. (1.3 m/s^{2}\nb. (2.5 m/s^{2}\nc. (- 1.2 m/s^{2}\nd. (0.1 m/s^{2})

a stuntman drives a car with a mass of 1600 kg on a drawbridge. the car accelerates with a constant force of 8000 n. while he is driving, the drawbridge is raised to an incline of 30°. what is the cars new acceleration on this incline, ignoring the force due to air resistance? (recall that (g = 9.8 m/s^{2}))\nforce accelerating car\n30°\nweight\na. (1.3 m/s^{2}\nb. (2.5 m/s^{2}\nc. (- 1.2 m/s^{2}\nd. (0.1 m/s^{2})

Answer

Explanation:

Step1: Calculate the component of the gravitational force along the incline

The gravitational - force component along the incline is $F_{g\parallel}=mg\sin\theta$, where $m = 1600$ kg, $g = 9.8$ m/s², and $\theta = 30^{\circ}$. $F_{g\parallel}=1600\times9.8\times\sin30^{\circ}=1600\times9.8\times0.5 = 7840$ N.

Step2: Use Newton's second - law to find the acceleration

Newton's second - law is $F_{net}=ma$. The net force acting on the car along the incline is $F_{net}=F - F_{g\parallel}$, where $F = 8000$ N. $F_{net}=8000 - 7840=160$ N. Since $F_{net}=ma$, then $a=\frac{F_{net}}{m}$. Substituting $F_{net}=160$ N and $m = 1600$ kg, we get $a=\frac{160}{1600}=0.1$ m/s².

Answer:

D. $0.1$ m/s²