Toboggan Physics Problem: Grade 12 Assignment Help | Work & Friction

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The discussion revolves around a Grade 12 physics assignment involving a child pulling a toboggan up a hill and calculating work done against gravity. To find the work required to pull the toboggan at a constant velocity, the force applied by the child is calculated using the combined mass and gravitational acceleration, leading to a work value of 18,530.5 Joules for a 25.7-degree incline. A similar calculation for a different angle, such as 30 degrees, yields 16,245.6 Joules. When sliding down, the total work done is zero, as the forces balance out, with the gravitational force acting downward. Understanding these calculations is essential for mastering the concepts of work and friction in physics.
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I'm in grade 12 physics and I'm having a lot of trouble so far. I have an assignment due tomorrow but I can't get this one hard question. Any help would be appreciated.

A 25.6kg child pulls a 4.81kg toboggan up a hill inclined at 25.7 degrees to the horizontal. The vertical height of the hill is 27.3 meters. Friction is negligible.

a) Determine how much work the child must do on the toboggan to pull it at a constant velocity up the hill.

b) This part of the question just asks your to repeat a) with a different angle.

c) The child now slides down the hill on the toboggan. Determine the total work on the child and the toboggan during the slide.

Thank you for your time.
 
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W = mgh

add the masses and multiply times gravity and hieght.

The total work will be zero, up is positive and down is negative.

Nautica
 


Hello! I completely understand your struggle with this physics problem. It can be challenging to apply concepts and equations to real-life scenarios, but with practice and some guidance, you can definitely tackle it.

To solve this problem, we need to use the formula for work, which is W = Fd, where W is work, F is the force applied, and d is the distance traveled. We also need to consider the force of gravity and the angle of the hill.

a) To determine the work the child must do on the toboggan, we need to find the force applied by the child. This can be calculated using the formula F = mg, where m is the mass and g is the acceleration due to gravity (9.8 m/s^2). So, the force applied by the child is (25.6kg + 4.81kg)(9.8 m/s^2) = 297.6N. Now, we need to find the distance traveled, which is the vertical height of the hill (27.3m) divided by the sine of the angle (25.7 degrees). So, d = 27.3m / sin(25.7) = 62.3m. Finally, we can calculate the work done by the child: W = (297.6N)(62.3m) = 18,530.5 Joules.

b) To repeat this calculation with a different angle, we simply need to plug in the new angle into the formula. Let's say the new angle is 30 degrees. So, d = 27.3m / sin(30) = 54.6m. The work done by the child would then be W = (297.6N)(54.6m) = 16,245.6 Joules.

c) To determine the total work during the slide, we need to consider the force of gravity pulling the child and toboggan down the hill. The force of gravity can be calculated using the formula F = mg, where m is the combined mass of the child and toboggan (25.6kg + 4.81kg = 30.41kg) and g is still 9.8 m/s^2. So, F = (30.41)(9.8) = 297.6N. Now, the distance traveled during the slide is the same as in part a) and b
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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