- #1

Astronomer107

- 31

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So, basically, the most important thing in dealing with force is: F(net)= F(app)- F(res).

1. A race car has a mass of 710 kg. It starts from rest and travels 40.0 m in 3.0 seconds. The car is uniformly accelerated during the entire time. What net force is exerted on it?

So, acceleration = F(net)/m

V= d/t ; v = 40.0m/3.0s = 13.3 m/s

V(f)^2 = V(0)^2 + 2ad

(13.3)^2 = 0^2 + 2a (40.0)

176.9 = 80a

a= 2.21 m/s^2

Now, a= F(net)/ m

2.21 m/s^2 = F(net)/710

**F(net) = 1569.1 N**Is that correct?

2. A 225 kg crate is pushed horizontally with a force of 710 N. If the coefficient of friction is 0.20, calculate the acceleration of the crate.

F(net) = F(app)- F(res); F(app) = 710 N ; F(f)= F(n)M&; F(n) = mg ??

F(n) = 225(9.8) = 2205

F(f) = (0.20)(2205) = 441

F(net) = 710 N - 441 N = 269 N

A = F(net)/M

A = 269 N/ 225 kg

**A= 1.2 m/s^2**? I'm not sure I'm getting this...

52. A block lies on a smooth plane tilted at an angle of 22.0 degrees to the horizontal. (a) Determine the acceleration of teh block as it slides down the plane. (b) If the back starts from the rest 9.10m up the plane from its base, what will be the block's speed when it reaches the bottom of the incline? Ignore friction.

Since the verticle component of this is 9.8 m/s^2, 9.8/sin 22.2= 26.2 m/s^2. Thus, the acceleration of the box

is 26.2 m/s^2, but I don't understand how that can be. How can the acceleration down a slope be greater than gravitational acceleration?? I don't know how to do the next part. I feel like I have a very vague understanding of physics.

THANK YOU!