I was wondering if my answers are correct. Thanks in advance.

1. A boy pushes down on a car rolling horizontally down the road. The boy pushes down vertically with 10N of force as the car rolls 3.0m horizontally. What work is done?
my answer: 30 N•m (or would it be -30N•m?)

2. What is the speed of a 0.25kg ball if its kinetic energy is 15 J?
my answer: 11m/s

3. A ball is dropped off a high cliff. What quantity increases as the ball gets closer to the ground?
my answer: I'm not too sure. It's either kinetic energy or gravitational potential energy. I think kinetic energy

4. a 10.0 kg crate slides across the floor 13m against a force of friction of 7.5N. What is the thermal energy produced?
my answer: 98 J

5. In the process of being thrown, a 0.500 kg ball goes fromo rest to a speed of 19.9m/s over a distance of 0.75m. What is the force that was exerted on the ball?
my answer: 19 N

6. A 10.0kg crate initially at rest sitting on the floor with a static and kinetic coefficient of 0.20 is pushed with a force of 15N for 4.0s. What is the work done?
my answer: I'm not sure. I think it's 60 J

2. What is the speed of a 0.25kg ball if its kinetic energy is 15 J?

my answer: 11m/s
(1/2)(.25) * v^2 = 15
v = Yes, about 11 m/s

3. A ball is dropped off a high cliff. What quantity increases as the ball gets closer to the ground?
my answer: I'm not too sure. It's either kinetic energy or gravitational potential energy. I think kinetic energy

Yes, potential energy goes down, kinetic energy goes up with speed (1/2) m v^2

4. a 10.0 kg crate slides across the floor 13m against a force of friction of 7.5N. What is the thermal energy produced?
my answer: 98 J

Yes, 97.5 Joules

5. In the process of being thrown, a 0.500 kg ball goes fromo rest to a speed of 19.9m/s over a distance of 0.75m. What is the force that was exerted on the ball?
my answer: 19 N
F = change in momentum / change in time
= .5 (19.9-0)/t
t = distance / average speed = .75/10 = .075 seconds
so
F = .5 * 19 / .075 = 127 N

6. A 10.0kg crate initially at rest sitting on the floor with a static and kinetic coefficient of 0.20 is pushed with a force of 15N for 4.0s. What is the work done?
my answer: I'm not sure. I think it's 60 J
-------
weight = m g = 98.1 N
friction force = .2*98.1 = 19.6 N
total force = 15 -
LOL - another trick question
the crate DOES NOT MOVE
therefore NO WORK DONE

I WANT TO BE A PHYSISES

1. A boy pushes down on a car rolling horizontally down the road. The boy pushes down vertically with 10N of force as the car rolls 3.0m horizontally. What work is done?

my answer: 30 N•m (or would it be -30N•m?)

NO, No, No !!
Work is Force IN THE DIRECTION OF MOTION times the distance

Answer is ZERO !!!

thanks a lot!

You are welcome. Nasty questions !

1. The work done is actually zero because the force is applied vertically and the displacement is horizontal. The force is perpendicular to the displacement, so no work is done. But hey, at least that car got a good upper body workout!

2. Ah, the speed of the ball. If it's running late, you can always tell it to "kinetic energy up!" The correct answer is indeed 11m/s. Just make sure the ball doesn't get a speeding ticket!

3. As the ball gets closer to the ground, its gravitational potential energy decreases while its kinetic energy increases. So you were right on the money with kinetic energy! It's like the ball is saying, "I'm getting closer to the ground, kinetic energy, let's make some moves!"

4. The thermal energy produced in this case is actually 0 J. You may have slid off the answer there. The force of friction is perpendicular to the displacement, so no work is done and no thermal energy is produced. Time for that crate to cool off somewhere else!

5. The force exerted on the ball is actually 26.53 N. Maybe the ball was just having a tough day and needed that extra push. Time to put that force to use!

6. The work done on the crate is indeed 60 J. Looks like someone is gaining some momentum, at least in terms of work. Just make sure the crate doesn't lose its balance and fall into an existential crisis!

Remember, it's always good to double-check your answers and have a sense of humor when it comes to physics!

Great job on attempting the questions! Let's go through each question to check your answers.

1. To calculate work, you need to multiply force and displacement. In this case, the force is 10N (vertically downwards) and the displacement is 3.0m (horizontally). Since the force and displacement are perpendicular to each other, the work done would be zero. So, your answer should be 0 N•m.

2. The kinetic energy is given by the formula KE = (1/2)mv^2, where m is the mass and v is the velocity (speed). Rearranging the formula, we get v = sqrt(2KE/m). Plugging in the values, we have v = sqrt((2 * 15 J) / 0.25 kg), which gives v ≈ 10.95 m/s. So, your answer should be approximately 10.95 m/s.

3. As the ball gets closer to the ground, its gravitational potential energy decreases because it is losing height. So, the correct answer is gravitational potential energy.

4. The thermal energy produced can be found using the formula W = Fd, where W is the work done (thermal energy produced), F is the force of friction, and d is the distance. Plugging in the values, we have W = (7.5 N) * (13 m), which gives W = 97.5 J. So, your answer should be 97.5 J.

5. To calculate the force exerted on the ball, you can use Newton's second law, F = ma. In this case, the mass of the ball is 0.500 kg and the acceleration can be found using the formula a = (vf^2 - vi^2) / (2d), where vf is the final velocity, vi is the initial velocity (which is 0 m/s), and d is the distance. Plugging in the values, we have a = (19.9 m/s)^2 / (2 * 0.75 m), which gives a ≈139.2 m/s^2. Now we can calculate the force, F = (0.500 kg) * (139.2 m/s^2), which gives F ≈ 69.6 N. So, your answer should be approximately 69.6 N.

6. To calculate the work done, we need to multiply force and displacement. In this case, the force is 15N and the displacement is not given. However, since the crate is initially at rest and is being pushed for 4.0s, we can assume that the displacement is the product of the average velocity and time. The average velocity can be found using v = a * t, where a is the acceleration (calculated by dividing the force by the mass) and t is the time. Plugging in the values, we have v = (15 N) / (10 kg) * (4.0s), which gives v = 6 m/s. Now we can calculate the displacement, d = v * t, which gives d = (6 m/s) * (4.0s), which gives d = 24 m. Finally, we can calculate the work done, W = (15N) * (24m), which gives W = 360 J. So, your answer should be 360 J.

Overall, good job on your answers! Just a correction needed for question 1, where the work done would be zero.