Question

5. A race car has a mass of 710 kg. It starts from rest and travels 40.0m in 3.0s. The car is uniformly accelerated during the entire time. What net force is acting on the car? 6. Suppose that a 1000 kg car is traveling at 25 m/s (≈55 mph). Its brakes can apply a force of 5000N. What is the minimum distance required for the car to stop? 7. A 65 kg person dives into the water from the 10 m platform. a) What is her speed as she enters the water? b) She comes to a stop 2.0 m below the surface of the water. What net force was exerted on the swimmer? 8. During a head-on collision, a passenger in the front seat of a car accelerates from 13.3 m/s (≈ 30 miles/hour) to rest in 0.10 s. a) What is the acceleration of the passenger? b) The driver of the car holds out his arm to keep his 25 kg child (who is not wearing a seat belt) from smashing into the dashboard. What force must he exert on the child? c) What is the weight of the child? d) Convert these forces from N to pounds. (x 1 lb 4.45N ). What are the chances the driver will be able to stop the child?

Answer 1

5. 6319 N

First of all, we need to find the acceleration of the car, which can be found by using the equation

$S=\frac{1}{2}at^2$

where

S = 40.0 m is the distance travelled by the car

t = 3.0 s is the time taken

a is the acceleration

Solving for a, we find

$a=\frac{2S}{t^2}=\frac{2(40.0 m)}{(3.0 s)^2}=8.9 m/s^2$

So now since we know the mass of the car, m=710 kg, we can find the net force acting on the car:

$F=ma=(710 kg)(8.9 m/s^2)=6319 N$

6. 62.5 m

In this case, we know the breaking force applied on the car,

$F=-5000 N$

(with a negative sign since its direction is opposite to the car's motion)

and the mass of the car

$m=1000 kg$

so we can find its acceleration:

$a=\frac{F}{m}=\frac{-5000 N}{1000 kg}=-5 m/s^2$

So now we can find the minimum distance to stop by using the equation

$v^2-u^2 = 2ad$

where in this case we have

v = 0 m/s is the final speed

u = 25 m/s is the initial speed

a = -5 m/s^2 is the acceleration

d is the distance

solving for d,

$d=\frac{v^2-u^2}{2a}=\frac{0^2-(25 m/s)^2}{2(-5 m/s^2)}=62.5 m$

7a. 14 m/s

We can solve the problem by using the law of conservation of energy: in fact, the initial gravitational potential energy of the person is all converted into kinetic energy as she hits the water below

$mgh=\frac{1}{2}mv^2$

where

m = 65 kg is the mass of the person

g = 9.8 m/s^2

h = 10 m is the initial height of the diver

v is the final speed as she enters the water

Solving for v, we find

$v=\sqrt{2gh}=\sqrt{2(9.8 m/s^2)(10 m)}=14 m/s$

7b. -3185 N

We need to find the acceleration of the diver during the motion of 2.0 m below the water:

$v^2-u^2 = 2ad$

where

v = 0 is the final speed

u = 14 m/s is the initial speed as she enters the water

a is the acceleration

d = 2.0 m is the distance covered

Solving for a,

$a=\frac{v^2-u^2}{2d}=\frac{0^2-(14 m/s)^2}{2(2.0 m)}=-49 m/s^2$

And so now we can find the net force acting on the diver

$F=ma=(65 kg)(-49 m/s^2)=-3185 N$

8a. -133 m/s^2

The acceleration of the passenger is given by

$a=\frac{v-u}{t}$

where

v = 0 m/s is the final speed

u = 13.3 m/s is the initial speed

t = 0.10 s is the time interval

Solving for a, we find

$a=\frac{0-13.3 m/s}{0.10 s}=-133 m/s^2$

8b. 3325 N

The force that the driver must exert on the child is equal in magnitude to the force experienced by the child during the stop of the car, so

$F=ma$

where

m = 25 kg is the mass of the child

$a=-133 m/s^2$ is the acceleration

So, the magnitude of the force will be

$F=(25 kg)(133 m/s^2)=3325 N$

8c.  245 N

The weight of the child is given by

$W=mg$

where

m = 25 kg is the child's mass

g = 9.8 m/s^2 is the acceleration due to gravity

Solving the equation,

$W=(25 kg)(9.8 m/s^2)=245 N$

8d. 55.1 lb

Since we know that

$1 lb = 4.45 N$

We can find the weight in pounds by setting the following proportion

$1 lb : 4.45 N = x : 245 N$

Solving for x,

$x=\frac{(1 lb)(245 N)}{4.45 N}=55.1 lb$

8e. Chances are very low

The force that the driver should exert on the child is

F = 3325 N

This force is equivalent to the force required to lift an object of mass m:

$F=mg\\m=\frac{F}{g}=\frac{3325 N}{9.8 m/s^2}=339.3 kg$

So, it is equivalent to the force required to lift an object of 339.3 kg, which is quite a lot. therefore, the changes are very low.