A wet bar of soap (m = 150g) slides freely down a ramp 3.0 m l...

Question

A wet bar of soap (m = 150g) slides freely down a ramp 3.0 m long inclined at 8.5°. How long does it take to reach the bottom? How would this change if the soap's mass were 250 g?

Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. A 180 g damp sponge starting from rest slips smoothly down a 3.5 m slope at 9.5 degrees. Determine the time it takes for the sponge to reach the bottom. Does altering the sponges mass to 280 g affect this duration. So that's our end goal. So ultimately, we're trying to figure out how long it will take this sponge to slide down this slope, which is another way of saying incline for the. So literally, the sponge is located on this slope or this incline. And we're trying to figure out how long it will take this sponge to slide down this incline, this slope to the bottom. And we're trying to figure out that if the time will be affected if they increase the sponges mass to 280 g, and if it will affect the time duration, it takes to slide down the slope. Awesome. We're also given some multiple choice answers. Let's read them off to see what our final answer might be. A is 2.1 seconds time unaffected by mass change, B is 4.4 seconds. Time changes with mass C is 5.7 seconds time unaffected by mass change. And finally, D is 2.1 seconds. Time changes with mass. Awesome. So first off, let us note that on a frictionless slope, the only force causing acceleration along the slope, which is denoted as FX is the component of gravity that is parallel to the slope, which we can write as M multiplied by G where M is the mass multiplied by G, the gravity multiplied by sine of theta where theta is the angle. So once again, M is the mass of the sponge. G is the acceleration due to gravity and theta is the incline angle of the slope. So now we must recall and apply noon's second law which states that the net force on an object is equal to its mass times its acceleration, which can be written as the sum of FX is equal to M multiplied by A Awesome. So now we need to apply this law to our sponge and we can write that M multiplied by G multiplied by sine of theta is equal to M multiplied by A. So we can simplify this equation by canceling out the unit or the canceling out M. So we can write that A equals G multiplied by sine of theta. So from this equation, we can see that acceleration is only dependent on gravity and the slopes angle not the mass. So now we must recall and use the equation for motion at a constant acceleration which states that X minus X zero is equal to V zero multiplied by T plus one half. So one half multiplied by acceleration a multiplied by T squared where T is the time. So let us note that for the case of this problem, V zero is the initial velocity and V zero in this case is zero, since the sponge will start from rest, so it'll start moving from rest. So with that in mind, we can simplify the motion equation and rearrange it to isolate and sulfur T which is one of our final answers that we are asked to solve for. So when we do that, we will get that T is equal to the square root of two multiplied by X minus X zero all divided by A, the acceleration. So now we can substitute in all of our known variables. And sulfur T note that we need to substitute in. And let's make a side note of this in blue, we need to substitute in A equals G multiplied by sine of theta. And we also need to note that X minus X zero is equal to the slope length. OK. So with that in mind, we can plug in our known variables. So two is multiplied by X minus X zero, which the value is given to us as 3.5 m. In our problem itself is it says it slips smoothly down a 3.5 m slope divided by the acceleration A which is 9.8 m per second squared. Because that's the numerical value for the acceleration due to gravity multiplied by sine of 9.5 degrees. Bi we're given 9.5 degrees in our problem itself for the incline angle to slope angle. Awesome. So when we plug that into our calculator, we will find that T is equal to 2.1 seconds when we round to one decimal place. And that is our first answer. And now we need to solve for our second answer. The time it takes for the sponge to reach the bottom is approximately 2.1 seconds. And this time is unchanged regardless of the sponges change in mass because mass does not influence the acceleration, which means that the correct answer has to be the letter, a 2.1 seconds in time unaffected by mass change. Thank you so much for watching. Hopefully, that helped and I can't wait to see you in the next video. Bye.

A wet bar of soap (m = 150g) slides freely down a ramp 3.0 m long inclined at 8.5°. How long does it take to reach the bottom? How would this change if the soap's mass were 250 g?

Key Concepts

Inclined Plane

Newton's Second Law of Motion

Kinematic Equations

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