Was having trouble with these 3 physics questions and would...

1. Tom pushes a 10-kg box up a 40-degree frictionless incline 8 meters with a force of 120N. (A) What is the work done by Tom? (B) What is the work done by gravity? (C) What is the total work? 2. A tennis ball with a mass of 0.058kg is launched from a catapult with an initial speed of 6.8 m/s and an initial height of 1.3 meters. Assuming level ground, what is the final speed of the ball right before it strikes the ground? 3. A student pulls on a cart and applies a 20N force at an angle of 25 degrees above the horizontal to move a crate a distance of 10m across a rough floor that creates a frictional force of 2N. (A) Draw a FBD (B) How much work does the student do? (C) What is the total work done? (D) If the cart starts at rest, how fast is it going after 10m?

Transcript

00:01 Hi, here in this given problem out of the three different problems, i am solving the first two.

00:08 And in the first one, that is question number one.

00:12 Mass of the box which is being pushed by the term, that is m, is equal to 10 kilogram.

00:37 Angle of the ramp, that is 40 degree, and the force being applied by the term, that is, 120 newton.

00:47 This is the ramp at an angle of 40 degree the box being pushed up by the force f mass of the box m so its weight.

01:03 Mg acting vertically down having two components one of the component mg costs 40 degree perpendicular to this ramp another component m g sign 40 degree along the ramp in downward direction length means the distance moved over the ramp over the incline that is s is equal to 8 meter so in the first part of the problem work done by the term w t that is equal to the product of the force supplied by the term multiplied by the distance traveled.

01:58 So that is 120 multiplied by 8.

02:02 So this work done comes out to be equal to 960 joules.

02:09 And that is the answer for the first part of the problem.

02:12 Then in the second part of the problem we have to find work done by the gravity and that will be a negative work as the force due to gravity which is mg sign 40 degree and the displacement these are in opposite direction having 180 degree angle between them so that will be negative in nature so wg plugging in all the known values for the mass of the box 10 kilogram 9 .8 multiplied by sine 40 degree multiplied by 8 and this work is calculated to be equal to minus 500 103 .9 jules.

03:07 That is the answer for the second part of the problem.

03:12 Finally in the third part of the problem we have to find total work done over the box.

03:21 That will be equal to algebraic some of these two works.

03:25 Means this is 960 minus 503 .9 joules and it is calculated to be equal to 456 .1 jule answer for the third part of this problem then there is one more problem that is question number two in which mass of the tennis ball that is given as m is equal to 0 .058 kilogram it's initial speed after being launched from the catapult, v0, 6 .8 meter per second...

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