Physics help please? 10 pts. best!?

Question

Alright I've already attempted to solve these practice problems and I'm not thinking I did any right..just solve what you can and that'd be great! Thanks a million!

1. A tow truck exerts a force of 5974 N on a car, accelerating it at 3 m/s2. (We'll assume there is no friction on the car.)

Find the mass of the car in kg.

2. Let's say that the car from the previous problem has a mass of 995 kg. If the force of friction on the car is now 943 N and the tow truck pulls with a force of 6096 N, what will be the acceleration of the car, in m/s2?

3. Two perpendicular forces, one of 36 N directed upward and the second of 59 N directed to the right, act simultaneously on an object with a mass of 36 kg. What is the magnitude of the resultant acceleration, in m/s2, of the object?

4. A tow truck pulls a 1564 kg car with an acceleration of 4 m/s2.

What is the net force, in newtons, acting on the car?

5. You drag your heavy 5 kg backpack across the floor with a force of 9 N. If the frictional force acting on the backpack is 5 N, what is the acceleration, in m/s2, of the backpack?

6. A 15 g (mass) sparrow flying toward a bird feeder mistakes the pane of glass in a window for an opening and slams into it with a force of 2 N.
What is the bird’s acceleration?

7. In the previous question, let's say the bird experienced an acceleration of 63 m/s2 (regardless of what answer you really got).

How many “g’s” (units of gravitational acceleration) is the bird experiencing?

Answer 1

1. A tow truck exerts a force F of 5974 N on a car, accelerating it at a = 3 m/s2. (We'll assume there is no friction on the car.) Find the mass m of the car in kg.

mass m = Force F over acceleration a = 5974/3 = 1991.(3) kg



2. Let's say that the car from the previous problem has a mass m of 995 kg. If the force of friction Ff on the car is now 943 N and the tow truck pulls with a force F of 6096 N, what will be the acceleration of the car, in m/s2?

a = (F-Ff)/m = (6096-943)/995 = 5.18 m/sec^2



3. Two perpendicular forces, one of 36 N directed upward and the second of 59 N directed to the right, act simultaneously on an object with a mass m of 36 kg. What is the magnitude of the resultant acceleration a in m/s2, of the object?

resultant force Fr = √36^2+59^2 = 69.12 N

acceleration a = Fr/m = 69.12/36 = 1.92 m/sec^2



4. A tow truck pulls a 1564 kg car with an acceleration of 4 m/s2. What is the net force, in newtons, acting on the car?

force F = mass m times acceleration a = 1564*4 = 6256 N



5. You drag your heavy 5 kg backpack having mass m across the floor with a force F of 9 N. If the frictional force Ff acting on the backpack is 5 N, what is the acceleration a in m/s2, of the backpack?

acceleration a = (F-Ff)/m = (9-5)/5 = 4/5 = 0.8 m/sec^2



6. A 15 g (mass m) sparrow flying toward a bird feeder mistakes the pane of glass in a window for an opening and slams into it with a force F of 2 N. What is the bird’s acceleration a?

a = F/m = 2/0.015 = 2000/15 = 133.(3) m/sec^2



7. In the previous question, let's say the bird experienced an acceleration a of 63 m/s2 (regardless of what answer you really got). How many “g’s” (units of gravitational acceleration) is the bird experiencing?

g's = a/g = 63/9.8 = 6.43

Answer 2

1.) F=MA (remember that)

5974N=m (3m/s^2)

5947/3=m==》1991kg

2.) Again F=MA

6096N-943N=995kg (a)

a=5.17m/s^2

3.) (36N)^2 + (59N)^2=c^2 (if u draw it with a free body diagram you'll see that it's a right triangle)

C=69.12

Now f=MA

69.12=36kg (a)

A=1.95m/s^2





4.) F=MA

1564kg (4m/s^2)=6256N

5.) Okay for this one it's best to draw a free body diagram to visualize. Friction is acting against the pulling force therefore 9N-5N=Net force which is 4N. Then it's just f=MA

4N=5kg (a)==》a=.8m/s^2

6.) F=MA

2N=.015kg (a)

a=2/.015m/s^2 (punch this fraction in ur Calc 4 me plz)



7.) Remember g's is acceleration due to gravity so 1g=9.81m/s^2

So (63m/s^2)/(9.8m/s^2)=6.43g's

Answer 3

So in the first one, you have one horizontal force acting on the car, causing the acceleration. F = m*a, so (5974 N) = m*(3 m/s²). Recall that newton is the same as kg*m/s². so if you divide both sides by (3 m/s²), the remaining units are kilograms (which is indeed a mass unit). So the answer is (5974 kg*m/s²) / (3 m/s²) = 1991.33 kg.



In the second one, draw a diagram with the tow truck pushing to the right with 6096 N, and friction pushing to the left with 943 N. The net force to the right is 5153 N. With F = m*a, we have a = (5153 kg*m/s²) / (995 kg) = 5.179 m/s² to the right.



Find the hypotenuse of the right triangle with legs 36 N and 59 N. sqrt((36 N)² + (59 N)²) = 69.12 N. Divide that by 36 kg to find the magnitude of acceleration in that direction: 1.92 m/s²

Answer 4

This obviously is homework. By completing these, you will learn some basic physics.