physics problem,,can you help me GENIUS?

Question

WORK,POWER, AND ENERGY PROBLEM

1) john pushed a 20kg grocery cart from the canned goods section to the cashier. at a distance of 6m she applied a force of 100N at an angle of 15 degree above the horizontal. how much work was done by john on the grocery cart?

2) a 3kg ball is thrown with a velocity of 7.5m/s. what is the kenetic energy of the ball?

3) annie transferred a bundle of newspaper from the floor to the top of her 3m high cabinet. she exerted 15Joule of work what is the weight of the bundle of newspaper?

4)being a helpful student. you helped your physics teacher push the crate of new equipment for your school laboratory. you pushed the 50kg crate with a force of 240Newton while the frictional force of 96Newton opposed the crates motion. if you pushed the crate for 15m towards the laboratory room. how much work did you do? how much work was done by friction?how much work was done by the force of the gravity on the crate ? what is the total work done by all the force.?


5) a rock on top of a 300m cliff has a gravitational potential energy of 6000Joule. what is the weight of the rock?


IMPULSE AND MOMENTUM PROBLEM

1) carol, who is the 60kg, is driving her car at a speed of 25m/s. she sees a dog crossing the road and so she steps on the brake to avoid hitting it. her seatbelt. brings her body to a stop in 0.40second
(a) what force does the seat belt exert on her?
(b) if she is not trapped to her seatbelt. and the windshield stops her head in 0.001second.
what average force would the windshield exert on her?

2) kimbely is sitting on a luneta part bench watching the sunset when the childs ball rolls toward her. kimberly returns the ball to the child by hitting it with her 1.5kg bag at a speed of 15m/s. with what force does kimberly hit the ball when the time of contact last for 0.4second?



please help me genius...

please show me the correct solution
follow the format

GIVEN:
FIND?:
EQUATION:
SOLUTION:

THANK YOU :)

Answer 1

You know what is given and what is found and don't need anyone to retype it for you (do you?). I'll give you the equations.



1. W = 100N • cos15º • 6m = joules



2. Ek = ½mv² = ½•3kg•(7.5m/s)² = joules



3. W / d = F, so 15J / 3m = newtons



4. All of these are W=F•d questions (except, of course, gravity does no work).



5. Ep = mgh, so 6000J = (mg)•300m

Solve for mg.



Momentum:

1) F•(delta)t = m•(delta)v

a) F•0.4s = 60kg•25m/s

Solve for F.



b) F•0.001s = 60kg•25m/s

Solve for F.



2) Not enough information to solve, unless you assume that the handbag stops completely upon contact with the ball, in which case use the impulse equation above.

Answer 2

l'll do a couple,



1) First think about it. You know if john was pushing the cart parallel with the ground it would be full force of 100N. But since it is at a angle of 15 degree it will be a little less. Suppose he pushed straight down on cart with 100N, you know it wouldn't move at all because it would be a vertical force. So this force will be somewhere between 100N and 0N.



If you multiply 100N by sin15 degree you get about 26N and that is wrong because you know he didn't lose 74N of force just by pushing 15 degrees off angle. So try 100N * cos15 = 96.6N.

That sounds right. By pushing at 15 degree angle he pushes with force of 96.6N (lost 3.4N ). In other words, Newtons = force of gravity x mass. So by telling you the force was 100N the mass of the cart is already figured in that.



Equation of work: work = force x distance

work = 96.6N * 6 meter = 579.6 Joule of work

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2) Real easy, just plug in numbers:

KE = 1/2 mv^2

Kinetic Energy = 0.5 * 3kg * ( 7.5m/s) ^2

KE = 84.4 Joules

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3) When you pick something up in the air you give it potential energy (PE). Which means it has energy from force of gravity and can do work when it goes back to the ground.



PE = mass * gravity * height of mass = mgh



They gave you the work done as 15J. The work done could either be kinetic or potential energy. In this case use the PE:



PE = 15J

PE = mgh

15J = m * 9.8 m/s^2 * 3 meter

solve for m (mass):

mass = 15J / ( 9.8 * 3) = 0.51 kg , not very heavy newspaper

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4) The force student pushes with is opposite to the friction force holding it back.

Student force = 240N to the right

friction force = 96N to the left



Again , it ask for work.

Net force = force to the right - friction force to the left

Net force = 240N - 96N = 144N of force the student applies to crate



Work = force x distance

Work done by student = 144N * 15 meter = 2,160 Joules of work

Work done or lost to friction = 96N * 15 meter = 1,440 J



Work done by force of gravity on crate? Tricky question. Remember, work = force x distance.

force of gravity on crate = mg = 50kg * 9.8N = 490N. But since crate is on ground and gravity can't push it down any further, there is no distance the crate moved downward, therefore:



Work done by force of gravity on crate = 490N * 0 meter = Zero work done



Total work is just adding all the works together:



Work done by student = 2,160 J

Work done by friction = 1,440 J

Work done by gravity on crate = 0 J

Total work = 2,160J + 1,440J = 3,500 J

-----------------------------------------

5) PE = mgh

They give you that PE = 6,000J , and height of mass is 300 meter up. Gravity force is a is known at 9.8 Newons.



Plug in numbers:



PE = mgh

6,000J = mass * 9.8N * 300 meter

mass = 6,000J / (9.8N * 300m) = 2 kg

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Momentum one:



1) carol = 60kg

velocity = 25 m/s

stops in = 0.4 second



First , you have to figure acceleration. This is acceleration in reverse (deceleration) , but works same way:



use equation:

Velocity = initial velocity + acceleration * time

25 m/s = 0 m/s + acceleration * 0.4 second

acceleration = -62.5 m/s^2



In other words this car decelerated at a rate of 62.5 m/s^2 down to zero velocity



a) Now just take F = ma to figure force on the driver from belt

F = 60kg * 62.5 m/s^2 = 3,750N of force from seat belt on Carol



b) Gigantic difference here. Stopping in 0.001 second vs 0.4 second

same equation , different numbers



use equation:

Velocity = initial velocity + acceleration * time

25 m/s = 0 m/s + acceleration * 0.001 second

(de) acceleration = -23,148 m/s^2



F = ma

F = 60kg * -23,148 m/s^2 = 1,388,880 Newtons of force without seat belt



1,388,880N / 3,750N = 370 times more force felt on carol's face

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2) use this equation:

Impulse = Average force x time = mass x change in velocity



knowns:

initial velocity = 0 m/s

final velocity = 15 m/s

change in velocity = 15 m/s

change in time = 0.4 seconds

mass = 1.5 kg



plug in #:

Average force x time = mass x change in velocity

Avg force = mass x change in velocity / time

Avg force = (1.5kg * 15 m/s) / 0.4 s = 56.3 N



Crap, I did them all

Answer 3

Here comes the "Physics Genius" . . . . .!!! Introducing, Sir Isaac Newton !!! Any question folks?