3 Questions that have wracked my brain for two weeks

[frdave20]

[SOLVED] 3 Questions that have wracked my brain for two weeks

If anyone can help this physics novice out, I have been working on these problems for 2 weeks. I just started looking at an old Physics book, trying to brush up... so if you can help, here are the 4 questions:

1.) When you life a bowling ball with a force of 82 N, the ball accelerates upward with an acceleration, a. If you life with a force of 95 N, the balls acceleration is 4 x a. What is the weight of the ball? What is the acceleration, a?

2.) A 23kg suitcase is being pulled by a handle that is at an angle of 25 degrees above the horizontal. If the Normal force exerted on suitcase is 150 N, what is the force F applied to the handle?

3.) A shopper pushes a 8.9 kg shopping cart up a 13 degree incline. Find the horizontal force, F, needed to give the cart an acceleration of 1.35 m/s^2.

4.)On vacation, your 1300 kg car pulls a 540 kg trailer away from a stop light with an acceleration of 1.90 m/s^2. What is the net force exerted by the car on the trailer? what is the net force acting on the car?

I didn't realize that it would be this hard... I don't remember having this difficulty in Physics when I was younger...

If you can help, please leave a little blurb for an explanation...

thanx
frdave20

 

[HallsofIvy]

A "little blurb"? I don't know- I'm not good at "blurbing"!

1.) When you life a bowling ball with a force of 82 N, the ball accelerates upward with an acceleration, a. If you life with a force of 95 N, the balls acceleration is 4 x a. What is the weight of the ball? What is the acceleration, a?

You would have to support the bowling ball with a force equal to it's weight in order just to keep it from accelerating downward. Any force more than than causes the acceleration. Letting w be the weight of the ball and F the force upward, m the mass, and a the acceleration, F- w= ma. In the first case, F= 82 N so
82- w= ma. In the second, F= 95 and the acceleration is 4 times as great: 95- w= m(4a)= 4ma. You can treat ma as a single variable and solve the two equations for w.

2.) A 23kg suitcase is being pulled by a handle that is at an angle of 25 degrees above the horizontal. If the Normal force exerted on suitcase is 150 N, what is the force F applied to the handle?

Here, oddly, you don't need to know the mass of the suitcase. I take it that, by "Normal force", you mean vertical: normal to the ground. Break the force F into horizontal and vertical components:
horizontal is F cos[theta] and the vertical is F sin[theta]. Since you are told the vertical force is 150 N, you can find F from that.

3.) A shopper pushes a 8.9 kg shopping cart up a 13 degree incline. Find the horizontal force, F, needed to give the cart an acceleration of 1.35 m/s^2.

The acceleration, up the slope is 1.35 m/s^2. F= ma so there must be a NET force, up the slope of (8.9)(1.35) Newtons. Again, you need to separate the force into horizontal and vertical components. Once you've done that, you can determine the horizontal force.

4.)On vacation, your 1300 kg car pulls a 540 kg trailer away from a stop light with an acceleration of 1.90 m/s^2. What is the net force exerted by the car on the trailer? what is the net force acting on the car?

Again, force= mass times acceleration. Just calculate the force necessary to accelerate 540 kg (the trailer) at 1.9 m/s^2. That's the force the car needs to exert on the trailer. You could find the force necessary to accelerate the whole assembly (car and trailer) and then subtract the force the car exerts on the trailer to find the NET force, but there is no reason to add and then subtract: just find the force necessary to accelerate 1300 kg at 1.9 m/s^2.

 

[M98Ranger]

1. For the first question, we can use the formula F=ma to solve for the weight of the ball and the acceleration. Since we have two different forces (82N and 95N) and the resulting acceleration is 4 times greater, we can set up the following equation: 82N = (4 x a) x m, where m is the mass of the ball. We can also set up the equation 95N = a x m. From these two equations, we can solve for a and m by setting them equal to each other: 82N = (4 x a) x m = 95N = a x m. This gives us a value of a = 20.5 m/s^2 and m = 4.05 kg. To find the weight of the ball, we can use the formula W=mg, where g is the acceleration due to gravity (9.8 m/s^2). Therefore, the weight of the ball is 39.69 N.

2. For the second question, we can use the formula F=ma again, but this time we need to consider the forces in the y-direction (vertical) and the x-direction (horizontal). We can break down the force F into its components in the x-direction (F_x) and the y-direction (F_y). F_x will be the force that is parallel to the ground and F_y will be the force that is perpendicular to the ground. We can use the angle given (25 degrees) to solve for these components: F_x = F cos(25) and F_y = F sin(25). The normal force, N, is equal to the weight of the suitcase (mg) and we can solve for the weight using the formula W=mg. Therefore, we can set up the following equation: F_x = F cos(25) = W = mg = 150 N. Solving for F, we get a value of 176.47 N.

3. For the third question, we can use the formula F=ma again, but this time we need to consider the forces in the y-direction (vertical) and the x-direction (horizontal). We can break down the weight of the shopping cart into its components in the x-direction (W_x) and the y-direction (W_y). W_x will be the weight that is parallel to the incline and W_y will be the weight that is perpendicular