Two children tug on a toy of mass 2.0kg. Costa pulls with a force of 10.0N[W 30degrees S]. The toy slides with an acceleration of 2.5m/s^2 in the direction of the net force. (North and West are positive)
A) Determine the value of the frictional force acting on the toy.
b)Determine the coeffieicent of friction.

A) 23 joules

B)8

Explanation: trust me

vi = 18.5 m/s

vf = 46.1 m/s

t = 2.47 s

Find:

d = ??

a = ??

a = (Delta v)/t

a = (46.1 m/s - 18.5 m/s)/(2.47 s)

a = 11.2 m/s2

d = vi*t + 0.5*a*t2

d = (18.5 m/s)*(2.47 s)+ 0.5*(11.2 m/s2)*(2.47 s)2

d = 45.7 m + 34.1 m

d = 79.8 m

(Note: the d can also be calculated using the equation vf2 = vi2 + 2*a*d)

the velocity you are looking for is made up of two parts: horizontally velocity, vh, and its vertical velocity, vv. the actual velocity of the ball will be these two parts added together, and because velocity is a vector then they will need to be added vectorially.  

neglecting air resistance etc etc.,  

the horizontal velocity component will be constant (because there is no accelerating force in the horizontal direction.  

vh = 8 ms-1  

the vertical velocity component has an accelerating force: 'g'.  

using v = u +at where a = g = 9.81 ms-2.  

the vertical component of velocity, vv = 0 + gt = vv =9.81 x 2  

vv = 19.62 ms-1 . the zero appears because the virtical component of velocity at time t=0 is zero. remember, the ball was thrown horizontally.  

how fast it will be moving (it actual path (being a parabola )) will be the vector sum of these two.  

adding vectorally gives: -  

v = sqr root (8^2 + 19.62^2) = sqr root(64+384.9) = 21.18 or 21.2 ms-1

i'm on my way to history not a mistory.