Physics help, please!?

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coocooclock92

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I missed some time from class because I was moving, and I'm trying to catch up on my work.
I've been reading through my text book, but there are sill a few questions that I've either forgotten how to answer, or don't know how to go about answering.

I'll write a couple. This is a grade 11 U class.

-Motion Assignment

1. A force of 32 N [<--] is applied to a 8 Kg brick restig on a table top. u = 0.10. How long does it take the brick to reach a speed of 4.0 m/s?

WEP Unit

How would I determine my own kinetic energy (assuming I weigh 130 pounds - 58.9 KG) while running at a speed of 5.5 m/s.

-The Law of Conservation of Energy-

A 91-Kg kangaroo exerts enough force to acquire 2.7 kJ of kinetic energy in jumping straight upward.

a) Apply the law of conservation of energy to determine how high this agile marsupial jumps.

b) What is the magnitude of the kangaroo's maximum velocity?


I think that once I know how to answer these, I should be able to figure out any of the questions that come after. I appreciate any help. I'd really like to try and get the work I missed done before the holidays start. Four tests in the next two days, though, so I'm really going to have to stay focused.
 
1) Using F=ma,
32=8xa
a = 4
then use v=u+at
4=0.10+(4 x t)
t=10s

Kinetic E. = 0.5 x m x v^2
Try it yourself with that formula, m=mass, v=speed


For the next part, I really ain't sure because I haven't seen these type of qns in my physics lessons before. Hope it helped. (:


EDIT : Oh i know a)

Law of C.O.E means that P.E lost = K.E gained or vice versa

means mgh = 2.7kJ
-> 91 x g x h = 2.7kJ , take g to be acceleration of free fall as 10m/s^2

so h, height = 2.7/(91x10)
 
1) Using F=ma,
32=8xa
a = 4
then use v=u+at
4=0.10+(4 x t)
t=10s

Kinetic E. = 0.5 x m x v^2
Try it yourself with that formula, m=mass, v=speed


For the next part, I really ain't sure because I haven't seen these type of qns in my physics lessons before. Hope it helped. (:


EDIT : Oh i know a)

Law of C.O.E means that P.E lost = K.E gained or vice versa

means mgh = 2.7kJ
-> 91 x g x h = 2.7kJ , take g to be acceleration of free fall as 10m/s^2

so h, height = 2.7/(91x10)
 
1. The acceleration of the brick is a = F/m, where F is the total force. There are two forces acting on it, the applied force of 32 N and the frictional force Fr acting in the opposite direction. The total force F = 32 - Fr. The friction force Fr = µ*m*g, so

F = 32 - µ*m*g

and the acceleration is

a = F/m = 32/m - µ*g

once you have acceleration, the velocity is V = a*t, The time to reach velocity V is then V/a

2.

The kinetic energy must equal the potential energy at maximum height (where the velocity and kinetic energy are zero). Potential energy = m*g*h; h = KE/(m*g)

Kinetic energy is 0.5*m*v² which is maximum at the initial launch. So v = √[2*KE/m]
 
1. The acceleration of the brick is a = F/m, where F is the total force. There are two forces acting on it, the applied force of 32 N and the frictional force Fr acting in the opposite direction. The total force F = 32 - Fr. The friction force Fr = µ*m*g, so

F = 32 - µ*m*g

and the acceleration is

a = F/m = 32/m - µ*g

once you have acceleration, the velocity is V = a*t, The time to reach velocity V is then V/a

2.

The kinetic energy must equal the potential energy at maximum height (where the velocity and kinetic energy are zero). Potential energy = m*g*h; h = KE/(m*g)

Kinetic energy is 0.5*m*v² which is maximum at the initial launch. So v = √[2*KE/m]
 
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